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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.3
Find the integrals of the functions in Exercises 62 to 83.
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Question : 80 of 271
Marks: +1, -0
1sin⁡xcos⁡3x\frac{1}{\sin x \cos^3 x}
Solution:  
Let I=∫1sin⁡xcos⁡3xdxI=\int \frac{1}{\sin x \cos^3 x}dx
=∫sin⁡2x+cos⁡2xsin⁡xcos⁡3xdx=\int \frac{\sin^2 x + \cos^2 x}{\sin x \cos^3 x}dx
=∫(sin⁡2xsin⁡xcos⁡3x+cos⁡2xsin⁡xcos⁡3x)dx=\int \left( \frac{\sin^2 x}{\sin x \cos^3 x} + \frac{\cos^2 x}{\sin x \cos^3 x} \right)dx
=∫(sin⁡xcos⁡3x+cos⁡xsin⁡xcos⁡2x)dx=\int \left( \frac{\sin x}{\cos^3 x} + \frac{\cos x}{\sin x \cos^2 x} \right)dx
=∫(tan⁡xsec⁡2x+sec⁡2xtan⁡x)dx=\int \left( \tan x \sec^2 x + \frac{\sec^2 x}{\tan x} \right)dx
=∫(tan⁡x+1tan⁡x)sec⁡2xdx=\int \left( \tan x + \frac{1}{\tan x} \right) \sec^2 x dx
Put tan⁡x=t\tan x = t ⇒sec⁡2xdx=dt\Rightarrow \sec^2 x dx = dt
∴I=∫(t+1t)dt=t22+log⁡∣t∣+C\therefore I = \int \left( t + \frac{1}{t} \right) dt = \frac{t^2}{2} + \log |t| + C
=log⁡∣tan⁡x∣+12tan⁡2x+C=\log |\tan x| + \frac{1}{2} \tan^2 x + C
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