Definite Integration Question 48
Question 48
- Let $f(x)=7 \tan ^{8} x+7 \tan ^{6} x-3 \tan ^{4} x-3 \tan ^{2} x$ for all $x \in-\frac{\pi}{2}, \frac{\pi}{2}$. Then, the correct expression(s) is/are (a) $\int_{0}^{\pi / 4} x f(x) d x=\frac{1}{12}$ (b) $\int_{0}^{\pi / 4} f(x) d x=0$ (c) $\int_{0}^{\pi / 4} x f(x) d x=\frac{1}{6}$ (d) $\int_{0}^{\pi / 4} f(x) d x=1$
(2015 Adv.)
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Solution:
- Here, $f(x)=7 \tan ^{8} x+7 \tan ^{6} x-3 \tan ^{4} x-3 \tan ^{2} x$ for all $x \in \frac{-\pi}{2}, \frac{\pi}{2}$
$$ \begin{aligned} \therefore \quad f(x) & =7 \tan ^{6} x \sec ^{2} x-3 \tan ^{2} x \sec ^{2} x \ & =\left(7 \tan ^{6} x-3 \tan ^{2} x\right) \sec ^{2} x \end{aligned} $$
Now, $\int_{0}^{\pi / 4} x f(x) d x=\int_{0}^{\pi / 4} x\left(7 \tan ^{6} x-3 \tan ^{2} x\right) \sec ^{2} x d x$
$$ =\left[x\left(\tan ^{7} x-\tan ^{3} x\right)\right]_{0}^{\pi / 4} $$
$-\int_{0}^{\pi / 4} 1\left(\tan ^{7} x-\tan ^{3} x\right) d x$
$$ \begin{aligned} & =0-\int_{0}^{\pi / 4} \tan ^{3} x\left(\tan ^{4} x-1\right) d x \ & =-\int_{0}^{\pi / 4} \tan ^{3} x\left(\tan ^{2} x-1\right) \sec ^{2} x d x \end{aligned} $$
Put $\tan x=t \Rightarrow \sec ^{2} x d x=d t$
$\therefore \int_{0}^{\pi / 4} x f(x) d x=-\int_{0}^{1} t^{3}\left(t^{2}-1\right) d t$
$$ =\int_{0}^{1}\left(t^{3}-t^{5}\right) d t=\frac{t^{4}}{4}-{\frac{t^{5}}{5}}_{0}^{1}=\frac{1}{4}-\frac{1}{6}=\frac{1}{12} $$
Also, $\int_{0}^{\pi / 4} f(x) d x=\int_{0}^{\pi / 4}\left(7 \tan ^{6} x-3 \tan ^{2} x\right) \sec ^{2} x d x$
$$ =\int_{0}^{1}\left(7 t^{6}-3 t^{2}\right) d t=\left[t^{7}-t^{3}\right]_{0}^{1}=0 $$
