Let f(x) = 7 tan^8 x + 7 tan^6 x - 3 tan^4 x | vedclass

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(A) Given $f(x) = 7 	an^6 x (	an^2 x + 1) - 3 	an^2 x (	an^2 x + 1) = (7 	an^6 x - 3 	an^2 x) \sec^2 x$.First,evaluate $\int_0^{\pi/4} f(x) dx$:

Vedclass · Accepted Answer

$(A, B)$

Vedclass · Answer

(A) Given $f(x) = 7 	an^6 x (	an^2 x + 1) - 3 	an^2 x (	an^2 x + 1) = (7 	an^6 x - 3 	an^2 x) \sec^2 x$.First,evaluate $\int_0^{\pi/4} f(x) dx$:$\int_0^{\pi/4} (7 	an^6 x - 3 	an^2 x) \sec^2 x dx$. Let $u = 	an x$,then $du = \sec^2 x dx$. When $x=0, u=0$; when $x=\pi/4, u=1$.$= \int_0^1 (7u^6 - 3u^2) du = [u^7 - u^3]_0^1 = (1 - 1) - (0 - 0) = 0$. Thus,$(B)$ is correct.Next,evaluate $\int_0^{\pi/4} x f(x) dx$ using integration by parts:Let $I = \int_0^{\pi/4} x f(x) dx$. Let $g(x) = \int f(x) dx = 	an^7 x - 	an^3 x$.Using integration by parts: $\int_0^{\pi/4} x f(x) dx = [x g(x)]_0^{\pi/4} - \int_0^{\pi/4} g(x) dx$.$[x(	an^7 x - 	an^3 x)]_0^{\pi/4} = \frac{\pi}{4}(1 - 1) - 0 = 0$.So,$I = - \int_0^{\pi/4} (	an^7 x - 	an^3 x) dx = \int_0^{\pi/4} (	an^3 x - 	an^7 x) dx$.$= \int_0^{\pi/4} 	an^3 x (1 - 	an^4 x) dx = \int_0^{\pi/4} 	an^3 x (1 - 	an^2 x)(1 + 	an^2 x) dx$.$= \int_0^{\pi/4} (	an^3 x - 	an^5 x) \sec^2 x dx$. Let $u = 	an x$,$du = \sec^2 x dx$.$= \int_0^1 (u^3 - u^5) du = [\frac{u^4}{4} - \frac{u^6}{6}]_0^1 = \frac{1}{4} - \frac{1}{6} = \frac{3-2}{12} = \frac{1}{12}$. Thus,$(A)$ is correct.

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