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

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

Answers (1)

best_answer

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

Solution:

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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