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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise Very Short Answers Question 44

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Answer: \frac{-1}{4}\tan\left ( 7-4x \right )+c

Hint: You must know about the integral rule of trigonometric functions.

Given: \int \sec^{2}\left ( 7-4x \right )dx


\int \sec^{2}\left ( 7-4x \right )dx

Put (7-4x) = t   and differentiate both sides, \left [ \frac{d}{dx} ax=a\right ]

dx= \frac{-dt}{4}

\begin{aligned} &=\int \sec ^{2}(t) \frac{d t}{-4} \\ &=\frac{-1}{4} \int \sec ^{2}(t) d t \\ &=\frac{-1}{4} \tan (t)+c \\ &=\frac{-1}{4} \tan (7-4 x)+c\quad\quad\quad\quad\quad\quad \quad\left[\because \int \sec ^{2} x d x=\tan x+c\right] \end{aligned}

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