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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise Very Short Answers Question 28

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Answer: \frac{1}{2}\sec^{2}x+c

Hints: You must know about the integral rule of trigonometric functions

Given:  \int \frac{\sin x}{\cos^{3}x}dx


I=\int \frac{\sin x}{\cos^{3}x}dx

Let \cos x=t  differentiate both sides,  \left [ \frac{d}{dx}\cos x-\sin x \right ]



-\sin xdx=dt

I=\int \frac{-dt}{t^{3}}\\I=\int -t^{-3}dt\\I=\frac{-t^{-2}}{-3+1}+c\\=\frac{-t^{-2}}{-2}+c\\=\frac{1}{2t^{2}}+c\\=\frac{1}{2\cos^{2}x}+c \quad \quad \quad \quad \quad \quad \quad \quad \quad\left [ \int x^{n} dx =\frac{x^{n+1}}{n+1}\right ]\\=\frac{1}{2}\sec^{2}{ x}+c



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