Integration (LAQs)

Maths-2B | 6. Integration – LAQs:
Welcome to LAQs in Chapter 6: Integration. This page contains the most Important FAQs for Long Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.

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LAQ-1 : Evaluate the reduction formula for I_n=∫1sinⁿ xdx and hence find ∫1sin⁴ xdx

Given $$I_n = \int \sin^n x \, dx = \int \sin^{n-1}x(\sin x) \, dx$$

We take first function $$u = \sin^{n-1}x$$ and second function $$v = \sin x$$

$$\int v = -\cos x$$

From the integration by parts rule, we have

$$I_n = \sin^{n-1}x(-\cos x) – \int(n-1)\sin^{n-2}x \cos x(-\cos x) \, dx$$

$$= -\sin^{n-1}x\cos x + (n-1)\int \sin^{n-2}x\cos^2x \, dx$$

$$= -\sin^{n-1}x\cos x + (n-1)\int \sin^{n-2}x(1-\sin^2x) \, dx$$

$$= -\sin^{n-1}x\cos x + (n-1)\left[\int \sin^{n-2}x \, dx – \int \sin^n x \, dx\right]$$

$$= -\sin^{n-1}x\cos x + (n-1)I_{n-2} – (n-1)I_n$$

Rearranging, we get

$$nI_n = -\sin^{n-1}x\cos x + (n-1)I_{n-2} + I_n – I_n$$

Simplifying, we get

$$I_n = -\frac{\sin^{n-1}x\cos x}{n} + \frac{(n-1)}{n}I_{n-2}$$

Put n = 4 2 0 successively in equation (1), we get

$$I_4 = -\frac{\sin^3 x \cos x}{4} + \frac{3}{4}I_2$$

$$= -\frac{\sin^3 x \cos x}{4} + \frac{3}{4}\left[-\frac{\sin x \cos x}{2} + \frac{1}{2}I_0\right]$$

$$= -\frac{\sin^3 x \cos x}{4} – \frac{3}{8}\sin x \cos x + \frac{3}{8}I_0$$

$$= -\frac{\sin^3 x \cos x}{4} – \frac{3}{8}\sin x \cos x + \frac{3}{8}x + C$$

Given that $$I_0 = x$$ the final expression for $$I_4$$ is as follows:

$$I_4 = -\frac{\sin^3 x \cos x}{4} – \frac{3}{8}\sin x \cos x + \frac{3}{8}x + C$$

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LAQ-2 : Evaluate the reduction formula for I_n=∫1cosⁿ xdx and hence find ∫1cos⁴ xdx

Given $$I_n = \int \cos^n x \, dx = \int \cos^{n-1}x (\cos x) \, dx$$

We apply the integration by parts rule:

We take first function $$u = \cos^{n-1}x$$ and second function $$v = \cos x$$

$$\int v = \sin x$$

From the integration by parts rule, we have:

$$I_n = \cos^{n-1}x(\sin x) – \int (n-1) \cos^{n-2}x (-\sin x) \sin x \, dx$$

$$= \cos^{n-1}x(\sin x) + (n-1) \int \cos^{n-2}x (\sin^2 x) \, dx$$

$$= \cos^{n-1}x(\sin x) + (n-1) \int \cos^{n-2}x (1 – \cos^2 x) \, dx$$

$$= \cos^{n-1}x(\sin x) + (n-1) \int \cos^{n-2}x \, dx – (n-1) \int \cos^n x \, dx$$

$$= \cos^{n-1}x(\sin x) + (n-1)I_{n-2} – (n-1)I_n$$

Rearranging, we get:

$$I_n = \cos^{n-1}x(\sin x) + (n-1)I_{n-2} – nI_n + I_n$$

$$⇒ nI_n = \cos^{n-1}x(\sin x) + (n-1)I_{n-2} + I_n – I_n$$

$$⇒ I_n = \frac{\cos^{n-1}x(\sin x)}{n} + \frac{(n-1)}{n} I_{n-2}$$

Put n = 4, 2, 0 successively in equation (1), we get:

$$I_4 = \frac{\cos^3 x \sin x}{4} + \frac{3}{4}I_2$$

$$⇒ \frac{\cos^3 x \sin x}{4} + \frac{3}{4}\left[\frac{\cos x \sin x}{2} + \frac{1}{2}I_0\right]$$

$$= \frac{\cos^3 x \sin x}{4} + \frac{3}{8}\cos x \sin x + \frac{3}{8}I_0$$

$$= \frac{\cos^3 x \sin x}{4} + \frac{3}{8}\cos x \sin x + \frac{3}{8}x + C$$

Given that $$I_0 = x$$ the final expression for $$I_4$$ is as follows:

$$I_4 = \frac{\cos^3 x \sin x}{4} + \frac{3}{8}\cos x \sin x + \frac{3}{8}x + C$$

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LAQ-3 : Evaluate ∫1tanⁿ xdx, hence evaluate ∫1tan⁵ xdx, ∫1tan⁶ xdx

Let $$I_n = \int \tan^n x \, dx = \int (\tan^{n-2} x) \tan^2 x \, dx$$

$$= \int (\tan^{n-2} x) (\sec^2 x – 1) \, dx$$

$$= \int \tan^{n-2} x \sec^2 x \, dx – \int \tan^{n-2} x \, dx$$

$$= \frac{\tan^{n-1} x}{n-1} – I_{n-2}$$

Put n = 5, 3, 1 successively in (1), we get:

$$I_5 = \int \tan^5 x \, dx = \frac{\tan^4 x}{4} – I_3$$

$$= \frac{\tan^4 x}{4} – \left(\frac{\tan^2 x}{2} – I_1\right)$$

$$= \frac{\tan^4 x}{4} – \frac{\tan^2 x}{2} + \int \tan x \, dx$$

$$= \frac{\tan^4 x}{4} – \frac{\tan^2 x}{2} + \log |\sec x| + C$$

Put n = 6, 4, 2, 0 successively in (1), we get:

$$I_6 = \int \tan^6 x \, dx = \frac{\tan^5 x}{5} – I_4$$

$$= \frac{\tan^5 x}{5} – \left(\frac{\tan^3 x}{3} – I_2\right)$$

$$= \frac{\tan^5 x}{5} – \frac{\tan^3 x}{3} + \tan x – I_0$$

$$= \frac{\tan^5 x}{5} – \frac{\tan^3 x}{3} + \tan x – x + C$$

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LAQ-4 : Obtain reduction formula for I_n=∫1cotⁿ xdx, n being a positive integer, n≥2 and deduce the value of ∫1cot⁴ xdx

Let $$I_n = \int \cot^n x \, dx = \int (\cot^{n-2} x)\cot^2 x \, dx$$

$$= \int (\cot^{n-2} x)(\csc^2 x – 1) \, dx$$

$$= \int \cot^{n-2} x \cdot \csc^2 x \, dx – \int \cot^{n-2} x \, dx$$

$$= \frac{\cot^{n-1} x}{n-1} – I_{n-2}$$

Put n = 4, 2, 0 successively in (1) we get:

$$I_4 = \frac{\cot^3 x}{3} – I_2$$

$$= \frac{\cot^3 x}{3} – \left(\frac{\cot x}{1} – I_0\right)$$

$$= \frac{\cot^3 x}{3} – \cot x + \int \cot^0 x \, dx$$

$$= \frac{\cot^3 x}{3} – \cot x + \int dx$$

$$= \frac{\cot^3 x}{3} – \cot x + x + C$$

This results in the expression for $$I_4$$ as:

$$I_4 = -\frac{\cot^3 x}{3} + \cot x + x + C$$

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LAQ-5 : Obtain the reduction formula for I_n=∫secⁿ xdx. Hence find ∫sec⁵ xdx

Given $$I_n = \int \sec^n x \, dx = \int (\sec^{n-2} x)\sec^2 x \, dx$$

We take first function $$u = \sec^{n-2} x$$

Second function $$v = \sec^2 x$$

$$\int v = \tan x$$

From the integration by parts rule, we have:

$$I_n = (\sec^{n-2} x)\tan x – \int (n-2)(\sec^{n-3} x) \sec x \tan x (\tan x) \, dx$$

$$= (\sec^{n-2} x)(\tan x) – (n-2)\int (\sec^{n-2} x)\tan^2 x \, dx$$

$$= (\sec^{n-2} x)(\tan x) – (n-2)\int (\sec^{n-2} x)(\sec^2 x – 1) \, dx$$

$$= (\sec^{n-2} x)(\tan x) – (n-2)\int (\sec^n x – \sec^{n-2} x) \, dx$$

$$= (\sec^{n-2} x)(\tan x) – (n-2)I_n + (n-2)I_{n-2}$$

Rearranging:

$$I_n + (n-2)I_n = (\sec^{n-2} x)(\tan x) + (n-2)I_{n-2}$$

$$I_n[1 + (n-2)] = (\sec^{n-2} x)(\tan x) + (n-2)I_{n-2}$$

$$I_n(n-1) = (\sec^{n-2} x)(\tan x) + (n-2)I_{n-2}$$

$$I_n = \frac{(\sec^{n-2} x)(\tan x)}{n-1} + \frac{(n-2)}{n-1}I_{n-2}$$

Put n = 5, 3, 1 successively in (1) we get:

$$I_5 = \int \sec^5 x \, dx = \frac{\sec^3 x \tan x}{4} + \frac{3}{4}I_3$$

$$= \frac{\sec^3 x \tan x}{4} + \frac{3}{4}\left(\frac{\sec x \tan x}{2} + \frac{1}{2}I_1\right)$$

$$= \frac{\sec^3 x \tan x}{4} + \frac{3}{8}\sec x \tan x + \frac{3}{8} \int \sec x \, dx$$

$$= \frac{\sec^3 x \tan x}{4} + \frac{3}{8}\sec x \tan x + \frac{3}{8}\log|\sec x + \tan x| + C$$

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LAQ-6 : Obtain the reduction formula for I_n=∫cscⁿ xdx, and hence find ∫csc⁵ xdx

Given $$I_n = \int \csc^n x \, dx = \int \csc^{n-2} x \cdot \csc^2 x \, dx$$

We take first function $$u = \csc^{n-2} x$$

Second function $$v = \csc^2 x$$

$$\int v = -\cot x$$

From the integration by parts rule, we have:

$$I_n = -\csc^{n-2} x \cot x – \int (n-2) \csc^{n-3} x(-\csc x \cot x)(-\cot x) \, dx$$

$$= -\csc^{n-2} x \cot x – (n-2) \int \csc^{n-2} x \cot^2 x \, dx$$

$$= -\csc^{n-2} x \cot x – (n-2) \int \csc^{n-2} x (\csc^2 x – 1) \, dx$$

$$= -\csc^{n-2} x \cot x – (n-2) \int \csc^n x \, dx + (n-2) \int \csc^{n-2} x \, dx$$

Rearranging:

$$I_n = -\csc^{n-2} x \cot x – (n-2)I_n + (n-2)I_{n-2}$$

$$I_n + (n-2)I_n = -\csc^{n-2} x \cot x + (n-2)I_{n-2}$$

$$I_n(n-1) = -\csc^{n-2} x \cot x + (n-2)I_{n-2}$$

$$I_n = -\frac{\csc^{n-2} x \cot x}{n-1} + \frac{n-2}{n-1} I_{n-2}$$

Put n = 5, 3, 1 successively in (1), we get:

$$I_5 = \int \csc^5 x \, dx = -\frac{\csc^3 x \cot x}{4} + \frac{3}{4} I_3$$

$$= -\frac{\csc^3 x \cot x}{4} + \frac{3}{4}\left(-\frac{\csc x \cot x}{2} + \frac{1}{2}I_1\right)$$

$$= -\frac{\csc^3 x \cot x}{4} – \frac{3}{8} \csc x \cot x + \frac{3}{8} \int \csc x \, dx$$

$$= -\frac{\csc^3 x \cot x}{4} – \frac{3}{8} \csc x \cot x + \frac{3}{8} \log|\csc x – \cot x| + C$$

$$I_5 = -\frac{\csc^3 x \cot x}{4} – \frac{3}{8} \csc x \cot x + \frac{3}{8} \log| \tan \frac{x}{2} | + C$$

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LAQ-7 : Evaluate ∫dx/x²+x+1

$$x^2 + x + 1$$

$$x^2 + x + \left(\frac{1}{2}\right)^2 – \left(\frac{1}{2}\right)^2 + 1 = (x^2 + 2(x)\frac{1}{2} + \left(\frac{1}{2}\right)^2) + 1 – \frac{1}{4}$$

$$(x + \frac{1}{2})^2 + \frac{3}{4} = (x + \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2$$

$$\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}(\frac{x}{a}) + C$$

$$I = \int \frac{dx}{x^2 + x + 1} = \int \frac{dx}{(x + \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}$$

$$= \frac{1}{(\sqrt{3}/2)}\tan^{-1}\left(\frac{x + \frac{1}{2}}{(\sqrt{3}/2)}\right) + C = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) + C$$

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LAQ-8 : Evaluate ∫dx/4x²-4x-7

$$4x^2 – 4x – 7 = 4\left(x^2 – x – \frac{7}{4}\right) = 4\left(x^2 – x + \left(\frac{1}{2}\right)^2 – \left(\frac{1}{2}\right)^2 – \frac{7}{4}\right)$$

$$= 4\left(x^2 – 2\left(x\right)\frac{1}{2} + \left(\frac{1}{2}\right)^2 – \frac{1}{4} – \frac{7}{4}\right) = 4\left(\left(x – \frac{1}{2}\right)^2 – 2\right) = 4\left(\left(x – \frac{1}{2}\right)^2 – \left(\sqrt{2}\right)^2\right)$$

$$\frac{1}{x^2 – a^2}$$ $$\frac{1}{2a} \log \left|\frac{x – a}{x + a}\right| + C$$

$$I = \frac{1}{4} \int \frac{dx}{\left(x – \frac{1}{2}\right)^2 – \left(\sqrt{2}\right)^2}$$

$$= \frac{1}{8\sqrt{2}} \log \left|\frac{2x – 1 – 2\sqrt{2}}{2x – 1 + 2\sqrt{2}}\right| + C$$

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LAQ-9 : Evaluate ∫dx/(4cosx+3sinx)

Put $$\tan = \frac{x}{2} = t$$

$$\Rightarrow \sin(x) = \frac{2t}{1+t^2}; \cos(x) = \frac{1-t^2}{1+t^2}$$ and $$dx = \frac{2dt}{1+t^2}$$

$$I = \int\left(\frac{2dt}{1+t^2}\right)/\left[4\left(\frac{1-t^2}{1+t^2}\right)+3\left(\frac{2t}{1+t^2}\right)\right] = 2\int \frac{dt}{4-4t^2+6t} = – \frac{2}{4} \int \frac{dt}{t^2-\frac{3}{2}t-1}$$

$$= -\frac{1}{2}\int \frac{dt}{t^2-\frac{3}{2}t+\left(\frac{3}{4}\right)^2-\frac{9}{16}-1}$$

$$= -\frac{1}{2}\int \frac{dt}{\left(\frac{1-3/4}{2}\right)^2-\left(\frac{5}{4}\right)^2}$$

$$= -\frac{1}{2} \cdot \frac{1}{2}\left(\frac{5}{4}\right) \log\left|\frac{t-3/4-5/4}{t-3/4+5/4}\right|+c$$

$$= -\frac{1}{5}\log\left|\frac{t-2}{t+1/2}\right|+c$$

$$= \frac{1}{5} \log\left|\frac{2t+1}{2t-4}\right| + c$$

$$= \frac{1}{5} \log \left| \frac{2\tan\left(\frac{x}{2}\right)+1}{2\left(\tan\left(\frac{x}{2}\right)-2\right)} \right| + c$$

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LAQ-10 : Evaluate ∫1/(1+sinx+cosx) dx

Put $$\tan \frac{x}{2} = t$$ then $$\sin x = \frac{2t}{1+t^2}; \cos x = \frac{1-t^2}{1+t^2}$$ and $$dx = \frac{2dt}{1+t^2}$$

$$I = \int \frac{1}{1+\sin x + \cos x} dx = \int \frac{1}{1 + \frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2}} \left( \frac{2dt}{1+t^2} \right) = \int \frac{1}{\left( \frac{1+t^2}{1+t^2} \right) + \frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2}} \cdot \frac{2dt}{1+t^2}$$

$$= \int \frac{2dt}{2 + 2t} = \frac{2}{2} \int \frac{dt}{1+t} = \log |1+t| + c = \log |1 + \tan(x/2)| + c$$

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LAQ-11 : Evaluate ∫dx/(3cosx+4sinx+6)

Put $$\tan \frac{x}{2} = t$$ then $$\sin x = \frac{2t}{1+t^2}; \cos x = \frac{1-t^2}{1+t^2}$$ and $$dx = \frac{2dt}{1+t^2}$$

$$I = \int \frac{dx}{3\cos x + 4\sin x + 6} = \int \frac{1}{[3(1-t^2)/(1+t^2)+4(2t)/(1+t^2)+6]} \left( \frac{2dt}{1+t^2} \right) = \int \frac{1}{[3(1-t^2)+4(2t)+6(1+t^2)]/(1+t^2)} \left( \frac{2dt}{1+t^2} \right)$$

$$= \int \frac{2dt}{3(1-t^2)+4(2t)+6(1+t^2)}$$

$$= \int \frac{2dt}{3t^2+8t+9} = \frac{2}{3} \int \frac{dt}{(t+\frac{4}{3})^2+(\frac{\sqrt{11}}{3})^2}$$

$$= \frac{2}{\sqrt{11}} \tan^{-1}\left(\frac{3\tan \frac{x}{2} + 4}{\sqrt{11}}\right) + c$$

$$= \frac{2}{\sqrt{11}} \tan^{-1}\left(\frac{3\tan \frac{x}{2} + 4}{\sqrt{11}}\right) + c$$

$$3t^2 + 8t + 9 = 3\left(t^2 + \frac{8}{3}t + 3\right)$$

$$= 3\left(t^2 + \frac{2.4}{3}t + \left(\frac{4}{3}\right)^2 – \left(\frac{4}{3}\right)^2 + 3\right)$$

$$= 3\left(\left(t+\frac{4}{3}\right)^2+\frac{11}{9}\right) = 3\left(\left(t+\frac{4}{3}\right)^2+\left(\frac{\sqrt{11}}{3}\right)^2\right)$$

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LAQ-12 : Evaluate ∫9cosx-sinx/4sinx+5cosxdx

Let $$(9\cos x – \sin x)$$

$$= A \frac{d}{dx}(4\sin x + 5\cos x) + B(4\sin x + 5\cos x)$$

$$(9\cos x – \sin x)$$

$$= A(4\cos x – 5\sin x) + B(4\sin x + 5\cos x)$$

$$= \cos x(4A + 5B) + \sin x(-5A + 4B)$$

Equating the coefficients of cosx, we get

$$4A + 5B = 9 \quad …(1)$$

Equating the coefficients of sinx, we get

$$-5A + 4B = -1 \quad …(2)$$

$$(1) \times 4 \Rightarrow 16A + 20B = 36 \quad …(3)$$

$$(2) \times 5 \Rightarrow -25A + 20B = -5 \quad …(4)$$

$$(3) – (4) \Rightarrow 41A = 41 \Rightarrow A = 1$$

From (1), $$5B = 9 – 4A = 9 – 4(1) = 5$$

$$5B = 5 \Rightarrow B = 1$$

Putting A = 1, B = 1 in (i) we get Nr.

$$9\cos x – \sin x = 1 \frac{d}{dx}(4\sin x + 5\cos x) + 1(4\sin x + 5\cos x)$$

$$I = \int \frac{9\cos x – \sin x}{4\sin x + 5\cos x} dx$$

$$= \int \frac{1 \frac{d}{dx}(4\sin x + 5\cos x) + 1(4\sin x + 5\cos x)}{4\sin x + 5\cos x} dx$$

$$= \int \frac{d}{dx}(4\sin x + 5\cos x) \frac{1}{4\sin x + 5\cos x} dx + \int 1 dx$$

$$= \log |4\sin x + 5\cos x| + x + c$$

$$\int \frac{f'(x)}{f(x)} dx = \log |f(x)| + c$$

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LAQ-13 : Evaluate ∫2cosx+3sinx/4cosx+5sinx dx

Let $$2\cos x + 3\sin x$$

$$= A \frac{d}{dx}(4\cos x + 5\sin x) + B(4\cos x + 5\sin x)$$

$$2\cos x + 3\sin x$$

$$= A(-4\sin x + 5\cos x) + B(4\cos x + 5\sin x)$$

$$= \cos x(5A + 4B) + \sin x(-4A + 5B)$$

Equating the coefficients of cosx, we get

$$5A + 4B = 2…(1)$$

Equating the coefficients of sinx, we get

$$-4A + 5B = 3…(2)$$

$$(2) \times 5 \Rightarrow -20A + 25B = 15…(3)$$

$$(1) \times 4 \Rightarrow 20A + 16B = 8…(4)$$

$$(3) + (4) \Rightarrow 41B = 23 \Rightarrow B = 23/41$$

From (2) $$4A = 5B – 3 = 5(23/41) – 3$$

$$= 115 – 123/41 = -8/41$$

$$4A = -8/41 \Rightarrow A = -2/41$$

Putting A = -2/41 B = 23/41 in (i) we get Nr

$$2\cos x + 3\sin x = -2/41 \frac{d}{dx}(4\sin x + 5\cos x) + 23/41(4\sin x + 5\cos x)$$

$$I = \int \frac{2\cos x + 3\sin x}{4\cos x + 5\sin x} dx$$

$$= \int -\frac{2}{41} \frac{d}{dx}(4\cos x + 5\sin x) + \frac{23}{41}(4\cos x + 5\sin x)/4\cos x + 5\sin x dx$$

$$= -\frac{2}{41}\int \frac{d}{dx}(4\cos x + 5\sin x)/4\cos x + 5\sin x dx + \frac{23}{41}\int 1 dx$$

$$= -\frac{2}{41} \log|4\cos x + 5\sin x| + \frac{23}{41}x + c$$

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LAQ-14 : Evaluate ∫2sinx+3cosx+4/3 sin⁡x+4 cos⁡x+5 dx

Let $$(2\sin x + 3\cos x + 4) = A(3\sin x + 4\cos x + 5) + B \frac{d}{dx} (3\sin x + 4\cos x + 5) + C \quad \text{(i)}$$

Then we have: $$(2\sin x + 3\cos x + 4) = A(3\sin x + 4\cos x + 5) + B(3\cos x – 4\sin x) + C$$

This simplifies to: $$(2\sin x + 3\cos x + 4) = \sin x(3A – 4B) + \cos x(4A + 3B) + (5A + C)$$

Equating the coefficients of sinx, we get:

$$3A – 4B = 2 \quad \text{(1)}$$

Equating the coefficients of cosx, we get:

$$4A + 3B = 3 \quad \text{(2)}$$

Equating the constant terms we get:

$$5A + C = 4 \quad \text{(3)}$$

Multiply (1) by 4:

$$12A – 16B = 8 \quad \text{(4)}$$

Multiply (2) by 3:

$$12A + 9B = 9 \quad \text{(5)}$$

Subtract (4) from (5):

$$25B = 1 \quad \Rightarrow \quad B = \frac{1}{25}$$

From (1), we have:

$$3A = 2 + 4B = 2 + \frac{4}{25} = \frac{50}{25} + \frac{4}{25} = \frac{54}{25} \quad \Rightarrow \quad A = \frac{18}{25}$$

From (3), we find:

$$C = 4 – 5A = 4 – 5\left(\frac{18}{25}\right) = 4 – \frac{90}{25} = \frac{100}{25} – \frac{90}{25} = \frac{10}{25} \quad \Rightarrow \quad C = \frac{2}{5}$$

Putting $$A = \frac{18}{25} B = \frac{1}{25} C = \frac{2}{5}$$ in (i) we get:

$$(2\sin x + 3\cos x + 4) = \left(\frac{18}{25}\right)(3\sin x + 4\cos x + 5) + \left(\frac{1}{25}\right)\frac{d}{dx}(3\sin x + 4\cos x + 5) + \frac{2}{5}$$

Now for the integral I:

$$I = \int \frac{2\sin x + 3\cos x + 4}{3\sin x + 4\cos x + 5} dx$$

$$= \frac{18}{25}\int 1 dx + \frac{1}{25}\int \frac{d}{dx}(3\sin x + 4\cos x + 5) dx + \frac{2}{5}\int \frac{dx}{3\sin x + 4\cos x + 5}$$

$$= \frac{18}{25}x + \frac{1}{25}\log|3\sin x + 4\cos x + 5| + \frac{2}{5}\int \frac{dx}{3\sin x + 4\cos x + 5}$$

Finally, to find the last integral we use substitution:

$$\tan \frac{x}{2} = t \Rightarrow \sin x = \frac{2t}{1+t^2}, \cos x = \frac{1-t^2}{1+t^2}, dx = \frac{2}{1+t^2}dt$$

$$\int \frac{dx}{3\sin x + 4\cos x + 5} = \int \frac{2dt}{(3\cdot \frac{2t}{1+t^2}) + (4\cdot \frac{1-t^2}{1+t^2}) + 5(1+t^2)}$$

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LAQ-15 : Evaluate ∫cos⁡x+3 sin⁡x+7/cos⁡x+sin⁡x+1 dx

Let $$\cos x + 3\sin x + 7 = A \frac{d}{dx}(\cos x + \sin x + 1) + B(\cos x + \sin x + 1) + C$$

This simplifies to:

$$\cos x + 3\sin x + 7 = A(-\sin x + \cos x) + B(\cos x + \sin x + 1) + C \quad \text{(1)}$$

Which becomes:

$$\cos x + 3\sin x + 7 = \cos x(A + B) + \sin x(-A + B) + (B + C)$$

Equating the coefficients of cosx, we have:

$$A + B = 1 \quad \text{(1)}$$

Equating the coefficients of sinx, we have:

$$-A + B = 3 \quad \text{(2)}$$

Equating the constant terms, we have:

$$B + C = 7 \quad \text{(3)}$$

Now (1)+(2) implies:

$$2B = 4 \quad \Rightarrow \quad B = 2$$

From (1), we have:

$$A = 1 – B = 1 – 2 = -1$$

From (3), we have:

$$C = 7 – B = 7 – 2 = 5$$

Putting A = -1 B = 2 C = 5 into (1) we get:

$$\cos x + 3\sin x + 7 = -1(-\sin x + \cos x) + 2(\cos x + \sin x + 1) + 5$$

The integral I is:

$$I = \int \frac{\cos x + 3\sin x + 7}{\cos x + \sin x + 1} dx$$

This becomes:

$$I = -\int \frac{-\sin x + \cos x}{\cos x + \sin x + 1} dx + 2\int \frac{\cos x + \sin x + 1}{\cos x + \sin x + 1} dx + 5\int \frac{1}{\cos x + \sin x + 1} dx$$

$$= -\log|\cos x + \sin x + 1| + 2x + 5\int \frac{1}{\cos x + \sin x + 1} dx \quad \text{(II)}$$

Now, we find

$$I_1 = \int \frac{1}{\cos x + \sin x + 1} dx$$

Substitute $$\tan \frac{x}{2} = t$$

$$\sin x = \frac{2t}{1 + t^2}, \quad \cos x = \frac{1 – t^2}{1 + t^2}, \quad dx = \frac{2dt}{1 + t^2}$$

The integral I₁​ becomes:

$$I_1 = \int \frac{2dt}{(1 – t^2) + 2t + (1 + t^2)} = \int \frac{2dt}{2 + 2t} = \int \frac{dt}{1 + t}$$

$$= \log|1 + t| + C$$

$$= \log|1 + \tan \frac{x}{2}| + C$$

From (II), the integral I is:

$$I = -\log|\cos x + \sin x + 1| + 2x + 5\log|1 + \tan \frac{x}{2}| + C$$

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LAQ-16 : Evaluate ∫2x+5/√x²-2x+10 dx

Let $$2x + 5 = A \frac{d}{dx}(x^2 – 2x + 10) + B \quad \text{(i)}$$

Then we have:

$$2x + 5 = A(2x – 2) + B$$

$$= 2Ax – 2A + B$$

Equating the coefficient of x, we get:

$$2A = 2 \quad \Rightarrow \quad A = 1$$

Equating the constant terms, we get:

$$-2A + B = 5 \quad \Rightarrow \quad B = 5 + 2A = 5 + 2 \cdot 1 = 7 \quad \Rightarrow \quad B = 7$$

Putting A = 1, B = 7 in (i), we get:

$$2x + 5 = 1 \frac{d}{dx}(x^2 – 2x + 10) + 7$$

The integral I is:

$$I = \int \frac{2x + 5}{\sqrt{x^2 – 2x + 10}} dx = \int \frac{\frac{d}{dx}(x^2 – 2x + 10)}{\sqrt{x^2 – 2x + 10}} dx + 7\int \frac{dx}{\sqrt{x^2 – 2x + 10}}$$

This simplifies to:

$$I = 2\sqrt{x^2 – 2x + 10} + 7\int \frac{dx}{\sqrt{(x – 1)^2 + 3^2}}$$

$$= 2\sqrt{x^2 – 2x + 10} + 7\sinh^{-1}\left(\frac{x – 1}{3}\right) + C$$

Since:

$$x^2 – 2x + 10 = (x – 1)^2 + 3^2$$

And we know that:

$$\int \frac{dx}{\sqrt{x^2 + a^2}} = \sinh^{-1}\left(\frac{x}{a}\right)$$

Thus, the final integral I is:

$$I = 2\sqrt{x^2 – 2x + 10} + 7\sinh^{-1}\left(\frac{x – 1}{3}\right) + C$$

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LAQ-17 : Evaluate ∫x+1/x²+3x+12 dx

Let $$x + 1 = A \frac{d}{dx}(x^2 + 3x + 12) + B \quad \text{(i)}$$

Then, we have:

$$x + 1 = A(2x + 3) + B$$

$$\Rightarrow x + 1 = 2Ax + (3A + B)$$

Equating the coefficients of x we get:

$$2A = 1 \quad \Rightarrow \quad A = \frac{1}{2}$$

Equating the constant terms we get:

$$3A + B = 1 \quad \Rightarrow \quad B = 1 – 3A = 1 – \frac{3}{2} \quad \Rightarrow \quad B = -\frac{1}{2}$$

Putting $$A = \frac{1}{2} B = -\frac{1}{2}$$ in (i), we get:

$$x + 1 = \frac{1}{2} \frac{d}{dx}(x^2 + 3x + 12) – \frac{1}{2}$$

The integral I is:

$$I = \int \frac{x + 1}{x^2 + 3x + 12} dx$$

$$= \frac{1}{2}\int \frac{d}{dx}(x^2 + 3x + 12) dx – \frac{1}{2}\int \frac{dx}{x^2 + 3x + 12}$$

This simplifies to:

$$I = \frac{1}{2}\log|x^2 + 3x + 12| – \frac{1}{2}\int \frac{dx}{(x + \frac{3}{2})^2 + \left(\frac{\sqrt{39}}{2}\right)^2}$$

$$= \frac{1}{2}\log|x^2 + 3x + 12| – \frac{1}{2} \cdot \frac{2}{\sqrt{39}}\tan^{-1}\left(\frac{2x + 3}{\sqrt{39}}\right) + C$$

$$= \frac{1}{2}\log|x^2 + 3x + 12| – \frac{1}{\sqrt{39}}\tan^{-1}\left(\frac{2x + 3}{\sqrt{39}}\right) + C$$

Since:

$$x^2 + 3x + 12 = (x + \frac{3}{2})^2 + \frac{39}{4}$$

And we know that:

$$\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)$$

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LAQ-18 : Evaluate ∫(3x-2)√2x²-x+1 dx

Let $$3x – 2 = A \frac{d}{dx}(2x^2 – x + 1) + B \quad \text{(i)}$$

Then we have:

$$3x – 2 = A(4x – 1) + B$$

$$\Rightarrow 3x – 2 = 4Ax – A + B\Rightarrow 3x – 2 = 4Ax – A + B$$

Equating the coefficients of x, we get:

$$4A = 3 \quad \Rightarrow \quad A = \frac{3}{4}$$

Equating the constant terms, we get:

$$-A + B = -2 \quad \Rightarrow \quad B = -2 + A = -2 + \frac{3}{4} = -\frac{5}{4}$$

Putting $$A = \frac{3}{4} B = -\frac{5}{4}$$ in (i), we get:

$$3x – 2 = \frac{3}{4} \frac{d}{dx}(2x^2 – x + 1) – \frac{5}{4}$$

The integral I is:

$$I = \int (3x – 2)\sqrt{2x^2 – x + 1} dx = \frac{3}{4}\int \frac{d}{dx}(2x^2 – x + 1)\sqrt{2x^2 – x + 1} dx – \frac{5}{4}\int \sqrt{2x^2 – x + 1} dx$$

$$= \frac{1}{2}(2x^2 – x + 1)^{\frac{3}{2}} – \frac{5}{4}\int \sqrt{2x^2 – x + 1} dx \quad \text{(ii)}$$

Now, we find $$\int \sqrt{2x^2 – x + 1}dx$$

Completing the square for $$2x^2 – x + 1$$

$$2x^2 – x + 1 = 2\left(x^2 – \frac{x}{2} + \frac{1}{2}\right) = 2\left(x^2 – \frac{x}{2} + \frac{1}{16} – \frac{1}{16} + \frac{8}{16}\right) = 2\left(\left(x – \frac{1}{4}\right)^2 + \frac{7}{16}\right)$$

The integral becomes:

$$\int \sqrt{2x^2 – x + 1}dx = \sqrt{2} \int \sqrt{\left(x – \frac{1}{4}\right)^2 + \left(\frac{\sqrt{7}}{4}\right)^2} dx$$

Using a substitution for the integral of the form $$\int \sqrt{x^2 + a^2}dx$$ and the result is usually in terms of $$\sinh^{-1}\left(\frac{x}{a}\right)$$ or $$\ln|x + \sqrt{x^2 + a^2}|$$ we get

$$= \sqrt{2}\left[\frac{(x – \frac{1}{4})}{2}\sqrt{\left(x – \frac{1}{4}\right)^2 + \frac{7}{16}} + \frac{7}{4\sqrt{2}}\sinh^{-1}\left(\frac{4(x – \frac{1}{4})}{\sqrt{7}}\right)\right] + C$$

From (ii) the integral I becomes:

$$I = \frac{1}{2}(2x^2 – x + 1)^{\frac{3}{2}} – \frac{5\sqrt{2}}{4}\left[\frac{(x – \frac{1}{4})}{2}\sqrt{\left(x – \frac{1}{4}\right)^2 + \frac{7}{16}} + \frac{7}{4\sqrt{2}}\sinh^{-1}\left(\frac{4(x – \frac{1}{4})}{\sqrt{7}}\right)\right] + C$$

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LAQ-19 : Find ∫(6x+5)√6-2x²+x  dx

Let $$6x + 5 = A \frac{d}{dx}(6 – 2x^2 + x) + B \quad \text{(i)}$$

This simplifies to:

$$6x + 5 = A(-4x + 1) + B$$

$$\Rightarrow 6x + 5 = -4Ax + (A + B)$$

Equating coefficients of x we get:

$$-4A = 6 \quad \Rightarrow \quad A = -\frac{6}{4} = -\frac{3}{2}$$

Equating constant terms we get:

$$A + B = 5 \quad \Rightarrow \quad B = 5 – A = 5 + \frac{3}{2} = \frac{13}{2}$$

Putting $$A = -\frac{3}{2} B = \frac{13}{2}$$ in (i) we get:

$$6x + 5 = -\frac{3}{2}(6 – 2x^2 + x) + \frac{13}{2}$$

The integral I is:

$$I = \int (6x + 5)\sqrt{6 – 2x^2 + x} dx = -\frac{3}{2}\int \frac{d}{dx}(6 – 2x^2 + x)\sqrt{6 – 2x^2 + x}dx + \frac{13}{2}\int \sqrt{6 – 2x^2 + x}dx$$

$$= -\frac{3}{2} \cdot \frac{2}{3}(6 – 2x^2 + x)^{\frac{3}{2}} + \frac{13}{2}\int \sqrt{6 – 2x^2 + x}dx \quad \text{(ii)}$$

Now we find $$\int \sqrt{6 – 2x^2 + x}dx$$

Completing the square for $$6 – 2x^2 + x$$

$$6 – 2x^2 + x = -2\left(x^2 – \frac{x}{2} – 3\right) = -2\left(x^2 – \frac{x}{2} + \frac{1}{16} – \frac{1}{16} – 3\right) = -2\left[\left(x – \frac{1}{4}\right)^2 – \frac{49}{16}\right]$$

$$= 2\left[\frac{49}{16} – \left(x – \frac{1}{4}\right)^2\right] = 2\left[\left(\frac{7}{4}\right)^2 – \left(x – \frac{1}{4}\right)^2\right]$$

The integral becomes:

$$\int \sqrt{6 – 2x^2 + x}dx = \sqrt{2}\int \sqrt{\frac{49}{16} – \left(x – \frac{1}{4}\right)^2} dx$$

Using a substitution for the integral of the form $$\int \sqrt{a^2 – x^2}dx$$ which results in an expression involving $$\sin^{-1}\left(\frac{x}{a}\right)$$ we get

$$= \sqrt{2}\left[\frac{(x – \frac{1}{4})}{2}\sqrt{\frac{49}{16} – \left(x – \frac{1}{4}\right)^2} + \frac{49}{32}\sin^{-1}\left(\frac{4(x – \frac{1}{4})}{7}\right)\right] + C$$

From (ii) the integral I becomes:

$$..I = -\left(6 – 2x^2 + x\right)^{\frac{3}{2}} + \frac{13\sqrt{2}}{2}\left[\frac{(x – \frac{1}{4})}{2}\sqrt{\frac{49}{16} – \left(x – \frac{1}{4}\right)^2} + \frac{49}{32}\sin^{-1}\left(\frac{4(x – \frac{1}{4})}{7}\right)\right] + C$$

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LAQ-20 : Evaluate ∫√5-x/x-2 dx

To integrate $$\int \frac{\sqrt{5 – x}}{x – 2} dx,$$

we transform it as: $$\int \frac{\sqrt{(5 – x)^2}}{(x – 2)(5 – x)} dx$$

$$= \int \frac{5 – x}{\sqrt{(x – 2)(5 – x)}} dx = \int \frac{5 – x}{\sqrt{7x – 10 – x^2}} dx$$

Letting $$5 – x = A \frac{d}{dx}(7x – 10 – x^2) + B \quad \text{(i)}$$

$$5 – x = A(7 – 2x) + B = -2Ax + (7A + B)$$

Equating coefficients of x we get:

$$-2A = -1 \quad \Rightarrow \quad A = \frac{1}{2}$$

Equating constant terms we get:

$$7A + B = 5 \quad \Rightarrow \quad B = 5 – 7A = 5 – \frac{7}{2} = \frac{10}{2} – \frac{7}{2} = \frac{3}{2}$$

Putting $$A = \frac{1}{2} B = \frac{3}{2}$$ in (i), we get:

$$5 – x = \frac{1}{2} \frac{d}{dx}(7x – 10 – x^2) + \frac{3}{2}$$

$$I = \int \frac{5 – x}{\sqrt{7x – 10 – x^2}} dx = \frac{1}{2} \int \frac{d}{dx}(7x – 10 – x^2) \frac{1}{\sqrt{7x – 10 – x^2}} dx + \frac{3}{2} \int \frac{1}{\sqrt{7x – 10 – x^2}} dx$$

$$= \frac{1}{2} \cdot 2\sqrt{7x – 10 – x^2} + \frac{3}{2} \int \frac{dx}{\sqrt{-x^2 + 7x – 10}}$$

Now, let’s find $$\int \frac{dx}{\sqrt{-x^2 + 7x – 10}}$$

Completing the square:

$$-x^2 + 7x – 10 = -\left(x^2 – 7x + 10\right) = -\left[\left(x – \frac{7}{2}\right)^2 – \left(\frac{7}{2}\right)^2 + 10\right] = -\left[\left(x – \frac{7}{2}\right)^2 – \frac{49}{4} + \frac{40}{4}\right] = -\left[\left(x – \frac{7}{2}\right)^2 – \frac{9}{4}\right]$$

$$= \frac{9}{4} – \left(x – \frac{7}{2}\right)^2 = \left(\frac{3}{2}\right)^2 – \left(x – \frac{7}{2}\right)^2$$

$$\int \frac{dx}{\sqrt{\left(\frac{3}{2}\right)^2 – \left(x – \frac{7}{2}\right)^2}} = \sin^{-1}\left(\frac{2x – 7}{3}\right) + C$$

From (ii), the integral I is:

$$I = \sqrt{7x – 10 – x^2} + \frac{3}{2} \sin^{-1}\left(\frac{2x – 7}{3}\right) + C$$