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Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Very short answer type question 7

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Answer: 1-\frac{\pi}{4}

Hint: You must know the integration rules of trigonometric function with its limits

Given: \int_{0}^{\frac{\pi}{4}} \tan ^{2} x \; d x

Solution:  \int_{0}^{\frac{\pi}{4}} \tan ^{2} x \; d x

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

\begin{aligned} &=[\tan x]_{0}^{\frac{\pi}{4}}-[x]_{0}^{\frac{\pi}{4}} \\\\ &=\tan \frac{\pi}{4}-\tan 0-\frac{\pi}{4}+0 \end{aligned}

\begin{aligned} &=1-0-\frac{\pi}{4}+0 \\\\ &=1-\frac{\pi}{4} \end{aligned}


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