9 Most FAQ’s of Definite Integrals Chapter in Inter 2nd Year Maths-2B (TS/AP)

7 Marks

LAQ-1 : Evaluate ∫₀^π xsinx/1+cos² x dx

We know $$\int_{0}^{a} f(x)dx = \int_{0}^{a} f(a-x)dx$$

$$I = \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2x} dx$$

$$= \int_{0}^{\pi} \frac{(\pi – x)\sin(\pi – x)}{1 + \cos^2(\pi – x)} dx$$

$$= \int_{0}^{\pi} \frac{(\pi – x)\sin x}{1 + \cos^2 x} dx$$

$$= \int_{0}^{\pi} \frac{\pi \sin x}{1 + \cos^2x} dx – \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2x} dx$$

$$= \int_{0}^{\pi} \frac{\pi \sin x}{1 + \cos^2x} dx – I$$

$$\Rightarrow 2I = \int_{0}^{\pi} \frac{\pi \sin x}{1 + \cos^2 x} dx$$

$$\Rightarrow I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2x} dx$$

Put $$\cos x = t$$

$$\Rightarrow \sin x dx = -dt$$

Also $$x = 0 \Rightarrow t = \cos 0 = 1$$ and $$x = \pi \Rightarrow t = \cos \pi = -1$$

$$I = \frac{\pi}{2} \int_{1}^{-1} \frac{-dt}{1+t^2} = \frac{\pi}{2} \int_{-1}^{1} \frac{dt}{1+t^2} = \frac{\pi}{2} [\tan^{-1}t]_{-1}^{1}$$

$$= \frac{\pi}{2} [\tan^{-1}(1) – \tan^{-1}(-1)]$$

$$= \frac{\pi}{2} \left(\frac{\pi}{4} + \frac{\pi}{4}\right) = \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^2}{4}$$

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LAQ-2 : Evaluate ∫₀^π xsin³x/1+cos²x dx

We know $$\int_{0}^{a} f(x)dx = \int_{0}^{a} f(a – x)dx$$

$$I = \int_{0}^{\pi} \frac{x \sin^3 x}{1 + \cos^2 x} dx = \int_{0}^{\pi} \frac{(\pi – x) \sin^3(\pi – x)}{1 + \cos^2(\pi – x)} dx$$

$$= \int_{0}^{\pi} \frac{(\pi – x) \sin^3 x}{1 + \cos^2 x} dx = \pi \int_{0}^{\pi} \frac{\sin^3 x}{1 + \cos^2 x} dx – \int_{0}^{\pi} \frac{x \sin^3 x}{1 + \cos^2 x} dx$$

$$= \pi \int_{0}^{\pi} \frac{\sin^3 x}{1 + \cos^2 x} dx – I$$

$$\Rightarrow I + I = 2I = \pi \int_{0}^{\pi} \frac{\sin^3 x}{1 + \cos^2 x} dx$$

$$\Rightarrow I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin^3 x}{1 + \cos^2 x} dx$$

$$= \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin^2 x \sin x}{1 + \cos^2 x} dx$$

$$= \frac{\pi}{2} \int_{0}^{\pi} \frac{(1 – \cos^2 x) \sin x}{1 + \cos^2 x} dx$$

Put $$\cos x = t$$ $$\Rightarrow \sin x dx = dt$$

$$x = 0 \Rightarrow t = \cos 0 = 1$$ and $$x = \pi \Rightarrow t = \cos \pi = -1$$

$$I = \frac{\pi}{2} \int_{-1}^{1} \frac{-(1 – t^2)}{1 + t^2} dt$$

$$= \frac{\pi}{2} \int_{-1}^{1} \frac{-1 + t^2}{1 + t^2} dt$$

$$= \frac{\pi}{2} \int_{-1}^{1} \frac{t^2 + 1 – 2}{t^2 + 1} dt$$

$$= \frac{\pi}{2} \int_{-1}^{1} \left(1 – \frac{2}{1 + t^2}\right) dt$$

$$= \frac{\pi}{2} \left[ t – 2 \tan^{-1} t \right]_{-1}^{1}$$

$$= \frac{\pi}{2} \left[-1 – 2(-\frac{\pi}{4}) – 1 + 2(\frac{\pi}{4})\right]$$

$$= \frac{\pi}{2} \left[\frac{\pi}{2} + \frac{\pi}{2} – 2\right] = \frac{\pi}{2}[\pi – 2]$$

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LAQ-3 : Evaluate ∫₀^π xsinx/1+sinx dx

We know $$\int_{0}^{a} f(x)dx = \int_{0}^{a} f(a – x)dx$$

$$I = \int_{0}^{\pi} \frac{x \sin x}{1 + \sin x} dx = \int_{0}^{\pi} \frac{(\pi – x) \sin(\pi – x)}{1 + \sin(\pi – x)} dx = \int_{0}^{\pi} \frac{(\pi – x) \sin x}{1 + \sin x} dx$$

$$= \pi \int_{0}^{\pi} \frac{\sin x}{1 + \sin x} dx – \int_{0}^{\pi} \frac{x \sin x}{1 + \sin x} dx = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \sin x} dx – I$$

$$\Rightarrow I + I = 2I = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \sin x} dx = \pi \int_{0}^{\pi} 1 – \frac{1}{1 + \sin x} dx$$

$$= \pi \left[ \int_{0}^{\pi} 1 dx – \int_{0}^{\pi} \frac{1}{1 + \sin x} dx \right]$$

$$= \pi^2 – \pi \int_{0}^{\pi} \frac{1}{1 + \sin x} dx \quad \text{(1)}$$

Now $$\int_{0}^{\pi} \frac{1}{1 + \sin x} dx = \int_{0}^{\pi} \frac{1 – \sin x}{(1 + \sin x)(1 – \sin x)} dx$$

$$= \int_{0}^{\pi} \frac{1 – \sin x}{1 – \sin^2 x} dx = \int_{0}^{\pi} \frac{1 – \sin x}{\cos^2 x} dx$$

$$= \int_{0}^{\pi} (\sec^2 x – \sec x \tan x) dx = \left[ \tan x – \sec x \right]_{0}^{\pi}$$

Since $$\tan(\pi) = \tan(0) = 0$$ and $$\sec(\pi) = \sec(0) = 1$$

$$= (0 – 1) – (0 – 1) = -2 \quad \text{(2)}$$

Hence from (1) and (2):

$$2I = \pi^2 – 2\pi$$

$$\Rightarrow I = \frac{\pi^2}{2} – \pi$$

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LAQ-4 : Evaluate ∫₀^π∕2 sin² x/cosx+sinx dx

We know $$\int_{0}^{a} f(x)dx = \int_{0}^{a} f(a – x)dx$$

$$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 x}{\cos x + \sin x} dx \quad \text{(1)}$$

$$\Rightarrow I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^2 x}{\sin x + \cos x} dx \quad \text{(2)}$$

Adding (1) & (2), we get:

$$2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 x + \cos^2 x}{\sin x + \cos x} dx$$

$$\Rightarrow 2I = \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} dx$$

$$\Rightarrow I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} dx$$

$$I = \frac{1}{2\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{dx}{\sin x \cdot \frac{1}{\sqrt{2}} + \cos x \cdot \frac{1}{\sqrt{2}}}$$

$$= \frac{1}{2\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{dx}{\sin(x + \frac{\pi}{4})}$$

$$= \frac{1}{2\sqrt{2}} \log \left| \tan \left(\frac{x}{2} + \frac{\pi}{8} \right) \right| \bigg|_{0}^{\frac{\pi}{2}}$$

$$= \frac{1}{2\sqrt{2}} \left[ \log \left| \tan \left(\frac{\pi}{4} + \frac{\pi}{8} \right) \right| – \log \left| \tan \left(\frac{\pi}{8} \right) \right| \right]$$

$$= \frac{1}{2\sqrt{2}} \left[ \log \left| \tan \left(\frac{3\pi}{8} \right) \right| – \log \left| \tan \left(\frac{\pi}{8} \right) \right| \right]$$

Using properties of logarithms:

$$= \frac{1}{2\sqrt{2}} \log \left( \frac{\tan(3\pi/8)}{\tan(\pi/8)} \right)$$

Since $$\tan(3\pi/8) = \tan(135^\circ) = \sqrt{2} + 1$$ and $$\tan(\pi/8) = \tan(22.5^\circ) = \sqrt{2} – 1$$

$$= \frac{1}{2\sqrt{2}} \log \left( \frac{\sqrt{2} + 1}{\sqrt{2} – 1} \right)$$

Multiplying the argument of the log by $$\frac{\sqrt{2} + 1}{\sqrt{2} + 1}$$

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

$$= \frac{1}{2\sqrt{2}} \log \left( (\sqrt{2} + 1)^2 \right)$$

$$= \frac{1}{\sqrt{2}} \log(\sqrt{2} + 1)$$

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LAQ-5 : Show that ∫₀^π∕2 x/sinx+cosx dx = π/2√2 log⁡(√2+1)

We know $$\int_{0}^{a} f(x)dx = \int_{0}^{a} f(a – x)dx$$

$$I = \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x + \cos x} dx = \int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2} – x}{\sin\left(\frac{\pi}{2} – x\right) + \cos\left(\frac{\pi}{2} – x\right)} dx$$

$$= \int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2} – x}{\cos x + \sin x} dx = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} \frac{dx}{\sin x + \cos x} – \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x + \cos x} dx = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} \frac{dx}{\sin x + \cos x} – I$$

$$\Rightarrow I + I = 2I = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} \frac{dx}{\sin x + \cos x}$$

$$\Rightarrow I = \frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{dx}{\sin x + \cos x}$$

$$= \frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{dx}{\sqrt{2}\left(\sin x \frac{1}{\sqrt{2}} + \cos x \frac{1}{\sqrt{2}}\right)} = \frac{\pi}{4\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{dx}{\sin\left(x + \frac{\pi}{4}\right)}$$

$$= \frac{\pi}{4\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \csc\left(x + \frac{\pi}{4}\right)dx$$

Using the property $$\int \csc x dx = \log | \tan \frac{x}{2} | + C$$ the integral becomes:

$$= \frac{\pi}{4\sqrt{2}} \log\left|\tan\left(\frac{x}{2} + \frac{\pi}{8}\right)\right|_{0}^{\frac{\pi}{2}}$$

$$= \frac{\pi}{4\sqrt{2}} \left[ \log\left|\tan\left(\frac{\pi}{4} + \frac{\pi}{8}\right)\right| – \log\left|\tan\left(\frac{\pi}{8}\right)\right| \right]$$

$$= \frac{\pi}{4\sqrt{2}} \left[ \log\left(\sqrt{2} + 1\right) – \log\left(\sqrt{2} – 1\right) \right]$$

$$= \frac{\pi}{4\sqrt{2}} \log\left( \frac{\sqrt{2} + 1}{\sqrt{2} – 1} \right)$$

$$= \frac{\pi}{4\sqrt{2}} \log\left( (\sqrt{2} + 1)^2 \right)$$ since $$\frac{\sqrt{2} + 1}{\sqrt{2} – 1} = (\sqrt{2} + 1)^2$$

$$= \frac{\pi}{2\sqrt{2}} \log(\sqrt{2} + 1)$$

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LAQ-6 : Evaluate ∫₀^π∕4 log⁡(1+tan⁡x)dx

We know $$\int_{0}^{a} f(x)dx = \int_{0}^{a} f(a – x)dx$$

$$I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x)dx$$

$$= \int_{0}^{\frac{\pi}{4}} \log\left[1 + \tan\left(\frac{\pi}{4} – x\right)\right]dx$$

$$= \int_{0}^{\frac{\pi}{4}} \log\left[\frac{1 + \tan(\frac{\pi}{4}) – \tan x}{1 + \tan(\frac{\pi}{4})\tan x}\right]dx$$

$$= \int_{0}^{\frac{\pi}{4}} \log\left[\frac{2 – \tan x}{1 + \tan x}\right]dx$$

Since $$\tan(\frac{\pi}{4}) = 1$$ we simplify further:

$$= \int_{0}^{\frac{\pi}{4}} \log\left[\frac{2}{1 + \tan x}\right]dx$$

$$= \int_{0}^{\frac{\pi}{4}} \left[\log 2 – \log(1 + \tan x)\right]dx$$

$$= \log 2 \int_{0}^{\frac{\pi}{4}} 1dx – \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x)dx$$

$$= \log 2 \left[x\right]_{0}^{\frac{\pi}{4}} – I$$

$$= \left(\log 2\right)\left(\frac{\pi}{4}\right) – I$$

$$I + I = \left(\log 2\right)\left(\frac{\pi}{4}\right)$$

$$2I = \left(\frac{\pi}{4}\right)\log 2$$

$$I = \left(\frac{\pi}{8}\right)\log 2$$

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LAQ-7 : Evaluate ∫₀¹ log⁡(1+x)/1+x² dx

Put $$x = \tan \theta$$

$$\Rightarrow dx = \sec^2 \theta d\theta$$

Also $$x = 0 \Rightarrow \theta = 0 x = 1 \Rightarrow \theta = \frac{\pi}{4}$$

And $$1 + x^2 = 1 + \tan^2 \theta = \sec^2 \theta$$

$$\int_{0}^{1} \frac{\log(1 + x)}{1 + x^2} dx = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan \theta) d\theta$$

$$I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan \theta) d\theta$$

$$= \int_{0}^{\frac{\pi}{4}} \log \left[ 1 + \tan \left( \frac{\pi}{4} – \theta \right) \right] d\theta$$

$$= \int_{0}^{\frac{\pi}{4}} \log \left[ 1 + \frac{1 – \tan \theta}{1 + \tan \theta} \right] d\theta$$

$$= \int_{0}^{\frac{\pi}{4}} \log \left[ \frac{(1 + \tan \theta) + (1 – \tan \theta)}{1 + \tan \theta} \right] d\theta$$

$$= \int_{0}^{\frac{\pi}{4}} \log \left( \frac{2}{1 + \tan \theta} \right) d\theta$$

$$= \int_{0}^{\frac{\pi}{4}} \left[ \log 2 – \log(1 + \tan \theta) \right] d\theta$$

$$= \log 2 \int_{0}^{\frac{\pi}{4}} 1 d\theta – \int_{0}^{\frac{\pi}{4}} \log(1 + \tan \theta) d\theta = \log 2 \left[ \theta \right]_{0}^{\frac{\pi}{4}} – I$$

$$\Rightarrow I + I = \log 2 \left( \frac{\pi}{4} \right)$$

$$\Rightarrow 2I = \left( \frac{\pi}{4} \right) \log 2$$

$$\Rightarrow I = \left( \frac{\pi}{8} \right) \log 2$$

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LAQ-8 : Evaluate ∫_a^b √((x-a)(b-x)) dx

$$\int_{a}^{b} \sqrt{(x-a)(b-x)} dx = \int_{a}^{b} \sqrt{-x^2 + (a + b)x – ab} dx = \int_{a}^{b} \sqrt{-(x^2 – (a + b)x + ab)}dx$$

$$= \int_{a}^{b} \sqrt{-(x^2 – (a + b)x + \left(\frac{a + b}{2}\right)^2 – \left(\frac{a + b}{2}\right)^2 + ab)}dx$$

$$= \int_{a}^{b} \sqrt{-\left[\left(x – \frac{a + b}{2}\right)^2 – \left(\frac{b – a}{2}\right)^2\right]}dx = \int_{a}^{b} \sqrt{\left(\frac{b – a}{2}\right)^2 – \left(x – \frac{a + b}{2}\right)^2} dx$$

$$= \left[\frac{1}{2}\left(x – \frac{a + b}{2}\right)\sqrt{(x – a)(b – x)} + \frac{1}{2} \left(\frac{b – a}{2}\right)^2 \sin^{-1} \left(\frac{x – \frac{a + b}{2}}{\frac{b – a}{2}}\right)\right]_{a}^{b}$$

$$= 0 + \frac{(b – a)^2}{8}[\sin^{-1}(1) – \sin^{-1}(-1)] = \frac{(b – a)^2}{8} \left[\frac{\pi}{2} + \frac{\pi}{2}\right] = \frac{(b – a)^2}{8}\pi = \frac{\pi}{8} (b-a)^2$$

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LAQ-9 : Evaluate _∫0^π∕4 sinx+cosx/9+16sin2x dx

Here, we take the substitution $$\sin x – \cos x = t$$

Then $$(\cos x + \sin x)dx = dt$$

Now x = 0 $$\Rightarrow t = \sin 0 – \cos 0 = 0 – 1 = -1$$ and $$x = \frac{\pi}{4}$$

$$\Rightarrow t = \sin \frac{\pi}{4} – \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} – \frac{1}{\sqrt{2}} = 0$$

Also $$(\sin x – \cos x)^2 = t^2$$

$$\Rightarrow \sin^2 x + \cos^2 x – 2\sin x\cos x = t^2$$

$$\Rightarrow 1 – \sin 2x = t^2$$

$$\Rightarrow \sin 2x = 1 – t^2$$

$$9 + 16\sin^2 x = 9 + 16(1-t^2) = 9 + 16 – 16t^2 = 25 – 16t^2$$

$$I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{9 + 16\sin 2x} dx = \int_{-1}^{0} \frac{dt}{25 – 16t^2} = \int_{-1}^{0} \frac{dt}{5^2 – (4t)^2}$$

$$= \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{1}{5} \log \left| \frac{5 + 4t}{5 – 4t} \right| \Bigg|_{-1}^{0}$$

$$= \frac{1}{40} \left[ \log \left| \frac{5 + 0}{5 – 0} \right| – \log \left| \frac{5 – 4(-1)}{5 + 4} \right| \right] = \frac{1}{40} \left[ \log 1 – \log \frac{1}{9} \right]$$

$$= \frac{1}{40} \left[ 0 – (-\log 9) \right] = \frac{1}{40} \log 9 = \frac{2 \log 3}{40} = \frac{\log 3}{20}$$