Let for $f(x)=7 an^8x+7 an^6x-3 an^4x-3 an^2x$, $I_1=\int_0^{\pi/4}f(x)\,dx$ and $I_2=\int_0^{\pi/4}xf(x)\,dx$. Then $7I_1+12I_2$ is equal to :

Question

Let for $f(x)=7	an^8x+7	an^6x-3	an^4x-3	an^2x$, $I_1=\int_0^{\pi/4}f(x)\,dx$ and $I_2=\int_0^{\pi/4}xf(x)\,dx$. Then $7I_1+12I_2$ is equal to :

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $f(x)=\sec^2x\,(7	an^6x-3	an^2x)$, so $I_1=\int_0^1(7t^6-3t^2)\,dt=0$. Integrating $I_2$ by parts with antiderivative $	an^7x-	an^3x$: $I_2=-\int_0^1t^3(t^2-1)\,dt=\frac1{12}$. Hence $7I_1+12I_2=1$. Answer (2).

Q. Let for $f(x)=7\tan^8x+7\tan^6x-3\tan^4x-3\tan^2x$, $I_1=\int_0^{\pi/4}f(x)\,dx$ and $I_2=\int_0^{\pi/4}xf(x)\,dx$. Then $7I_1+12I_2$ is equal to :

Explanation

Reviewed by Dr. Abubacker

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