Indefinite Integration 1 Question 67
68. Evaluate $\int _{-\pi / 3}^{\pi / 3} \frac{\pi+4 x^{3}}{2-\cos |x|+\frac{\pi}{3}} d x$.
$(2004,4$ M)
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Solution:
- Let $I=\int _{-\pi / 3}^{\pi / 3} \frac{\pi d x}{2-\cos |x|+\frac{\pi}{3}}+4 \int _{-\pi / 3}^{\pi / 3} \frac{x^{3} d x}{2-\cos |x|+\frac{\pi}{3}}$
$$ \begin{gathered} \text { Using } \int _{-a}^{a} f(x) d x=\int _0^{a} f(x) d x, \quad f(-x)=f(x) \\ \therefore \quad I=2 \int _0^{\pi / 3} \frac{\pi d x}{2-\cos |x|+\frac{\pi}{3}}+0 \\ \because \frac{x^{3} d x}{2-\cos |x|+\frac{\pi}{3}} \text { is odd } \\ I=2 \pi \int _0^{\pi / 3} \frac{d x}{2-\cos (x+\pi / 3)} \end{gathered} $$
Put $\quad x+\frac{\pi}{3}=t \Rightarrow d x=d t$
$\therefore \quad I=2 \pi \int _{\pi / 3}^{2 \pi / 3} \frac{d t}{2-\cos t}=2 \pi \int _{\pi / 3}^{2 \pi / 3} \frac{\sec ^{2} \frac{t}{2} d t}{1+3 \tan ^{2} \frac{t}{2}}$
Put $\quad \tan \frac{t}{2}=u \quad \Rightarrow \quad \sec ^{2} \frac{t}{2} d t=2 d u$
$$ \begin{aligned} \Rightarrow & I=2 \pi \int _{1 / \sqrt{3}}^{\sqrt{3}} \frac{2 d u}{1+3 u^{2}}=\frac{4 \pi}{3}\left[\sqrt{3} \tan ^{-1} \sqrt{3} u\right] _{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \\ & =\frac{4 \pi}{\sqrt{3}}\left(\tan ^{-1} 3-\tan ^{-1} 1\right)=\frac{4 \pi}{\sqrt{3}} \tan ^{-1} \frac{1}{2} \\ \therefore & \int _{-\pi / 3}^{\pi / 3} \frac{\pi+4 x^{3}}{2-\cos |x|+\frac{\pi}{3}} d x=\frac{4 \pi}{\sqrt{3}} \tan ^{-1} \frac{1}{2} \end{aligned} $$
