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#### Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 60 Maths Textbook Solution.

Answer: $k=\frac{1}{2}$

Hint: Use indefinite formula then put the limit to solve this integral

Given: $\int_{0}^{k} \frac{1}{2+8 x^{2}} d x=\frac{\pi}{16}$

Solution: $\int_{0}^{k} \frac{1}{2+8 x^{2}} d x=\frac{\pi}{16}$

\begin{aligned} &\Rightarrow \int_{0}^{k} \frac{1}{8\left(x^{2}+\frac{2}{8}\right)} d x=\frac{\pi}{16} \\ &\Rightarrow \frac{1}{8} \int_{0}^{k} \frac{1}{x^{2}+\frac{1}{4}} d x=\frac{\pi}{16} \\ &\Rightarrow \frac{1}{8} \int_{0}^{k} \frac{1}{x^{2}+\left(\frac{1}{2}\right)^{2}} d x=\frac{\pi}{16} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right] \\ &\Rightarrow \frac{1}{8}\left[\frac{1}{\frac{1}{2}} \tan ^{-1}\left(\frac{x}{\frac{1}{2}}\right)\right]_{0}^{k}=\frac{\pi}{16} \end{aligned}

\begin{aligned} &\Rightarrow \frac{1}{8} \times 2\left[\tan ^{-1} 2 x\right]_{0}^{k}=\frac{\pi}{16} \\ &\Rightarrow \frac{1}{4}\left[\tan ^{-1} 2 k-0\right]=\frac{\pi}{16} \\ &\Rightarrow \tan ^{-1} 2 k=\frac{4 \pi}{16} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan \frac{\pi}{4}=1\right] \\ &\Rightarrow \tan ^{-1} 2 k=\frac{\pi}{4} \\ &\Rightarrow 2 k=\tan \frac{\pi}{4}=1 \\ &k=\frac{1}{2} \end{aligned}

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