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Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 28 Maths Textbook Solution.

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Answer:  \frac{\pi}{4}-\frac{2}{3}

Given:  \int_{0}^{\frac{\pi}{4}} \tan ^{4} x d x

Hint: Use trigonometric identities and then integrate by parts


\int_{0}^{\frac{\pi}{4}} \tan ^{4} x d x=\int_{0}^{\frac{\pi}{4}} \tan ^{2} x \tan ^{2} x d x

=\int_{0}^{\frac{\pi}{4}} \tan ^{2} x\left(\sec ^{2} x-1\right) d x \\

\begin{aligned} & &=\int_{0}^{\frac{\pi}{4}} \tan ^{2} x \sec ^{2} x d x-\int_{0}^{\frac{\pi}{4}}\left(\sec ^{2} x-1\right) d x \end{aligned}

\begin{aligned} &p u t \\ & \end{aligned}

\tan x=t \\   (Differentiate w.r.t to x)

\sec ^{2} x d x=d t

\begin{aligned} &=\left(\frac{t^{3}}{3}\right)_{0}^{\frac{\pi}{4}}-\left(1-\frac{\pi}{4}\right)=\frac{1}{3}-1+\frac{\pi}{4} \\ & \end{aligned}



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