29 Most VSAQ’s of Integration Chapter in Inter 2nd Year Maths-2B (TS/AP)

2 Marks

VSAQ-1 : Evaluate ∫e^x(sinx+cosx)dx

Formula:

$$\int e^x(f(x) + f'(x))dx = e^x f(x) + c$$

Here

$$f(x) = \sin x$$

$$\Rightarrow f'(x) = \cos x$$

$$\int e^x(\sin x + \cos x)dx = e^x \sin x + c$$

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VSAQ-2 : Evaluate ∫e^x(secx+secxtanx)dx

Formula:

$$\int e^x(f(x) + f'(x))dx = e^x f(x) + c$$

Here

$$f(x) = \sec x$$

$$\Rightarrow f'(x) = \sec x \tan x$$

$$\int e^x(\sec x + \sec x \tan x)dx = e^x \sec x + c$$

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VSAQ-3 : Evaluate ∫e^x(tanx+log⁡ secx)dx

Here

$$f(x) = \log \sec x$$

$$\Rightarrow f'(x) = \tan x$$

$$\int e^x(\log \sec x + \tan x)dx = e^x \log \sec x + c$$

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VSAQ-4 : Evaluate ∫e^x (1 + tan² x + tanx)dx

$$I = \int e^x[(1 + \tan^2 x) + \tan x] dx$$

$$= \int e^x(\sec^2 x + \tan x) dx$$

Here

$$f(x) = \tan x$$

$$\Rightarrow f'(x) = \sec^2 x$$

$$\int e^x(\tan x + \sec^2 x) dx = e^x \tan x + c$$

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VSAQ-5 : Evaluate ∫e^x (1+xlogx/x)dx

Here

$$f(x) = \log x$$

$$\Rightarrow f'(x) = \frac{1}{x}$$

$$\int e^x\left(1 + \frac{x \log x}{x}\right)dx$$

$$= \int e^x\left(\frac{1}{x} + \log x\right)dx = e^x \log x + c$$

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VSAQ-6 : Evaluate ∫xe^x/(x+1)²  dx\

$$I = \int e^x \frac{x e^x}{(x + 1)^2} dx = \int \left[x + 1 – \frac{1}{(x + 1)^2}\right]e^x dx$$

$$= \int e^x \left[\frac{1}{x + 1} – \frac{1}{(x + 1)^2}\right] dx$$

$$= e^x \left(\frac{1}{x + 1}\right) + c = \frac{e^x}{x + 1} + c$$

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VSAQ-7 : Evaluate ∫(1/1-x²+1/1+x²)dx

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

$$= \tanh^{-1}x + \tan^{-1}x + C$$

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VSAQ-8 : Evaluate ∫dx/(x+1)(x+2)

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

$$= \int \left( \frac{1}{x + 1} – \frac{1}{x + 2} \right) dx = \log|x + 1| – \log|x + 2| + C$$

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VSAQ-9 : Evaluate ∫sec² x.csc² xdx

$$\int \sec^2 x \cdot \csc^2 x dx = \int \frac{1}{\cos^2 x \sin^2 x} dx$$

$$= \int \frac{\sin^2 x + \cos^2 x}{\cos^2 x \sin^2 x} dx$$

$$= \int \left( \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \right) dx$$

$$= \int \sec^2 x dx + \int \csc^2 x dx$$

$$= \tan x – \cot x + C$$

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VSAQ-10 : Evaluate ∫1/(coshx+sinhx) dx

$$\int \frac{1}{\cosh x + \sinh x} dx$$

$$= \int \frac{\cosh^2 x – \sinh^2 x}{\cosh x + \sinh x} dx$$

$$= \int \frac{(\cosh x – \sinh x)(\cosh x + \sinh x)}{\cosh x + \sinh x} dx$$

$$= \int (\cosh x – \sinh x) dx$$

$$= \sinh x – \cosh x + C$$

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VSAQ-11 : Evaluate ∫1/1+cox dx

$$\int \frac{1}{1 + \cos x} dx = \int \frac{1 – \cos x}{(1 + \cos x)(1 – \cos x)} dx$$

$$= \int \frac{(1 – \cos x)}{1 – \cos^2 x} dx = \int \frac{(1 – \cos x)}{\sin^2 x} dx$$

$$= \int \left(\frac{1}{\sin^2 x} – \frac{\cos x}{\sin^2 x}\right) dx$$

$$= \int \csc^2 x dx – \int \csc x \cot x dx$$

$$= -\cot x + \csc x + C$$

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VSAQ-12 : Evaluate ∫1+cos² x/1-cos2x dx

$$\int \frac{1 + \cos^2 x}{1 – \cos^2 x} dx = \int \frac{1 + \cos^2 x}{2\sin^2 x} dx$$

$$= \frac{1}{2}\int \left(\frac{1}{\sin^2 x} + \frac{\cos^2 x}{\sin^2 x}\right)dx$$

$$= \frac{1}{2}\int (\csc^2 x + \cot^2 x) dx$$

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

$$= \frac{1}{2}\int (2\csc^2 x – 1) dx$$

$$= \int \csc^2 x dx – \frac{1}{2}\int dx$$

$$= -\cot x – \frac{1}{2}x + c$$

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VSAQ-13 : Evaluate ∫sin⁡(logx)/x dx

Put $$\log x = t$$

$$\Rightarrow \frac{1}{x} dx = dt$$

$$I = \int \frac{\sin(\log x)}{x} dx = \int \sin t dt$$

$$= -\cos t + C = -\cos(\log x) + C$$

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VSAQ-14 : Evaluate ∫cot⁡(logx)/x dx

Put $$\log x = t$$

$$\Rightarrow \frac{1}{x} dx = dt$$

$$I = \int \frac{\cot(\log x)}{x} dx = \int \cot t dt$$

$$= \log |\sin t| + C = \log |\sin(\log x)| + C$$

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VSAQ-15 : Evaluate ∫1/xlogx dx

Put $$\log x = t$$

$$\Rightarrow \frac{1}{x} dx = dt$$

$$\int \frac{1}{x \log x} dx = \int \frac{dt}{t} = \log|t| + C = \log(\log x) + C$$

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VSAQ-16 : Evaluate ∫1/xlogx[log⁡(logx)] dx

Put $$\log(\log x) = t$$

$$\Rightarrow \frac{1}{\log x} \cdot \frac{1}{x} dx = dt$$

$$I = \int \frac{dt}{t} = \log|t| + C = \log(\log(\log x)) + C$$

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VSAQ-17 : Find ∫log⁡(1+x)/1+x dx

Put $$\log(1+x) = t$$

$$\Rightarrow \frac{1}{1+x} dx = dt$$

$$I = \int \frac{\log(1+x)}{1+x} dx = \int t dt = \frac{t^2}{2} + C$$

$$= \frac{(\log(1+x))^2}{2} + C$$

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VSAQ-18 : Find ∫(logx)²/x dx

Put $$\log x = t$$

$$\Rightarrow \frac{1}{x} dx = dt$$

$$\int \frac{(\log x)^2}{x} dx = \int t^2 dt = \frac{t^3}{3} + C$$

$$= \frac{1}{3}(\log x)^3 + C$$

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VSAQ-19 : Evaluate ∫sin⁡(Tan^-1x)/1+x² dx

Put $$\tan^{-1}x = t$$

$$\Rightarrow \frac{dx}{1+x^2} = dt$$

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

$$= -\cos(\tan^{-1}x) + C$$

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VSAQ-20 : Evaluate ∫1/√sin^-1x √1-x² dx

Put $$\sin^{-1}x = t$$

$$\Rightarrow \frac{1}{\sqrt{1-x^2}} dx = dt$$

$$I = \int \frac{dt}{\sqrt{t}} = 2\sqrt{t} + C = 2\sqrt{\sin^{-1}x} + C$$

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VSAQ-21 : Evaluate ∫e^x (1+x)/(cos²(xe^x) dx

Put $$x e^x = t$$

$$\Rightarrow \left[x e^x + e^x(1)\right]dx = dt$$

$$\Rightarrow e^x(1 + x)dx = dt$$

$$\int \frac{e^x(1 + x)}{\cos^2(x e^x)} dx = \int \frac{dt}{\cos^2 t}$$

$$= \int \sec^2 t dt = \tan t + C = \tan(x e^x) + C$$

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VSAQ-22 : Evaluate ∫e^x sine^x dx

Put $$e^x = t$$

$$\Rightarrow e^x dx = dt$$

$$I = \int \sin t dt = -\cos t + C$$

$$= -\cos(e^x) + C$$

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VSAQ-23 : Evaluate ∫cos³x sin⁡ x dx

Put $$\cos x = t$$

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

$$I = -\int t^3 dt = -\frac{t^4}{4} + C$$

$$= -\frac{\cos^4 x}{4} + C$$

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VSAQ-24 : Evaluate ∫sin⁴ x/cos⁶ x dx

$$\int \frac{\sin^4 x}{\cos^6 x} dx = \int \frac{\sin^4 x}{\cos^4 x} \cdot \frac{1}{\cos^2 x} dx$$

$$= \int \tan^4 x \sec^2 x dx$$

Put $$\tan x = t$$

$$\Rightarrow \sec^2 x dx = dt$$

$$I = \int t^4 dt = \frac{t^5}{5} + C = \frac{\tan^5 x}{5} + C$$

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VSAQ-25 : Evaluate ∫x⁸/1+x¹⁸ dx

Put $$x^9 = t$$

$$\Rightarrow 9x^8 dx = dt$$

$$\Rightarrow x^8 dx = \frac{1}{9} dt$$

$$\int \frac{x^8 dx}{1+x^{18}} = \int \frac{x^8 dx}{1+(x^9)^2} = \frac{1}{9} \int \frac{dt}{1+t^2} = \frac{1}{9} \tan^{-1}t + C = \frac{1}{9} \tan^{-1}(x^9) + C$$

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VSAQ-26 : Evaluate ∫dx/1+e^x

$$\int \frac{dx}{1+e^x} = \int \left(1 + e^x – \frac{e^x}{1 + e^x}\right)dx = \int \left(1 – \frac{e^x}{1+e^x}\right)dx$$

$$= x – \log(1 + e^x) + C$$

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VSAQ-27 : Evaluate ∫logx/x²  dx

From the “By parts rule”, we have

$$\int \frac{\log x}{x^2} dx = \left(\log x\right)\left(-\frac{1}{x}\right) + \int \frac{1}{x} \cdot \frac{1}{x} dx = -\frac{1}{x} \log x – \frac{1}{x} + C$$

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VSAQ-28 : Evaluate ∫e^x sinx dx

Let $$u = \sin x$$ and $$v = e^x$$

From the “By parts rule”, we have:

$$I = \int \sin x \cdot e^x dx = \sin x \cdot e^x – \int \cos x \cdot e^x dx$$

$$= \sin x \cdot e^x – \left[\cos x \cdot e^x – \int (-\sin x) \cdot e^x dx\right] = \sin x \cdot e^x – \cos x \cdot e^x – I$$

$$\Rightarrow 2I = (\sin x – \cos x) \cdot e^x$$

$$I = \frac{e^x}{2}(\sin x – \cos x) + C$$

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VSAQ-29 : Evaluate ∫cos√x dx

Put $$\sqrt{x} = t$$

$$\Rightarrow x = t^2$$

$$\Rightarrow dx = 2t dt$$

$$I = 2\int t \cdot \cos t \, dt = 2[t \sin t – \int 1 \cdot \sin t \, dt]$$

$$= 2t \sin t + 2 \cos t + C = 2\sqrt{x} \sin \sqrt{x} + 2\cos \sqrt{x} + C$$