SATHEE: Definite Integration Question 48

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 $x \in - \frac{π}{2}, \frac{π}{2}$ . Then, the correct expression(s) is/are (a) $\int_{0}^{π / 4} x f (x) d x = \frac{1}{12}$ (b) $\int_{0}^{π / 4} f (x) d x = 0$ (c) $\int_{0}^{π / 4} x f (x) d x = \frac{1}{6}$ (d) $\int_{0}^{π / 4} f (x) d x = 1$

(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 $x \in \frac{- π}{2}, \frac{π}{2}$

$\begin{aligned} ∴ f (x) & = 7 \tan^{6} x \sec^{2} x - 3 \tan^{2} x \sec^{2} x & = (7 \tan^{6} x - 3 \tan^{2} x) \sec^{2} x \end{aligned}$

Now, $\int_{0}^{π / 4} x f (x) d x = \int_{0}^{π / 4} x (7 \tan^{6} x - 3 \tan^{2} x) \sec^{2} x d x$

$= {[x (\tan^{7} x - \tan^{3} x)]}_{0}^{π / 4}$

$- \int_{0}^{π / 4} 1 (\tan^{7} x - \tan^{3} x) d x$

$\begin{aligned} = 0 - \int_{0}^{π / 4} \tan^{3} x (\tan^{4} x - 1) d x & = - \int_{0}^{π / 4} \tan^{3} x (\tan^{2} x - 1) \sec^{2} x d x \end{aligned}$

Put $\tan x = t \Rightarrow \sec^{2} x d x = d t$

$∴ \int_{0}^{π / 4} x f (x) d x = - \int_{0}^{1} t^{3} (t^{2} - 1) d t$

$= \int_{0}^{1} (t^{3} - t^{5}) d t = \frac{t^{4}}{4} - {\frac{t^{5}}{5}}_{0}^{1} = \frac{1}{4} - \frac{1}{6} = \frac{1}{12}$

Also, $\int_{0}^{π / 4} f (x) d x = \int_{0}^{π / 4} (7 \tan^{6} x - 3 \tan^{2} x) \sec^{2} x d x$

$= \int_{0}^{1} (7 t^{6} - 3 t^{2}) d t = {[t^{7} - t^{3}]}_{0}^{1} = 0$

Definite Integration Question 48

Question 48

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Definite Integration Question 48

Question 48

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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