Indefinite Integration 1 Question 2
3. The integral $\int _{\pi / 6}^{\pi / 3} \sec ^{2 / 3} x \operatorname{cosec}^{4 / 3} x d x$ is equal to
(a) $3^{5 / 6}-3^{2 / 3}$
(b) $3^{7 / 6}-3^{5 / 6}$
(c) $3^{5 / 3}-3^{1 / 3}$
(d) $3^{4 / 3}-3^{1 / 3}$
(2019 Main, 10 April II)
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Solution:
- Let $I=\int _{\pi / 6}^{\pi / 3} \sec ^{2 / 3} x \operatorname{cosec}^{4 / 3} x d x$
$$ \begin{aligned} & =\int _{\pi / 6}^{\pi / 3} \frac{1}{\cos ^{2 / 3} x \sin ^{4 / 3} x} d x \\ & =\int _{\pi / 6}^{\pi / 3} \frac{\sec ^{2} x}{(\tan x)^{4 / 3}} d x \end{aligned} $$
[multiplying and dividing the denominator by $\cos ^{4 / 3} x$ ] Put, $\tan x=t$, upper limit, at $x=\pi / 3 \Rightarrow t=\sqrt{3}$ and lower limit, at $x=\pi / 6 \Rightarrow t=1 / \sqrt{3}$
and $\sec ^{2} x d x=d t$
$$ \text { So, } \quad \begin{aligned} I & =\int _{1 / \sqrt{3}}^{\sqrt{3}} \frac{d t}{t^{4 / 3}}=\frac{t^{-1 / 3}}{-1 / 3} \\ & =-3 \frac{1}{3^{1 / \sqrt{3}}}-3^{1 / 6} \\ & =3 \cdot 3^{1 / 6}-3 \cdot 3^{-1 / 6} \\ & =3^{7 / 6}-3^{5 / 6} \end{aligned} $$
