Let f(x) = 7^8x+7^6x-3^4x-3^2x, and let I_1=_0^/4f(x),dx, I_2=_0^/4xf(x),dx.

Question

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

Accepted Answer

Factor: $f(x) = (7\tan^6x-3\tan^2x)\sec^2x$. Then

$$I_1 = \left[\tan^7x-\tan^3x\right]_0^{\pi/4} = 1-1 = 0$$

For $I_2$, integrate by parts with $u=x$, $dv = f(x)dx = d(\tan^7x-\tan^3x)$:

$$I_2 = \big[x(\tan^7x-\tan^3x)\big]_0^{\pi/4} - \int_0^{\pi/4}(\tan^7x-\tan^3x)\,dx$$

The boundary term is $0$. Writing $\tan^7x-\tan^3x = \tan^3x(\tan^2x-1)(\tan^2x+1) = (\tan^5x-\tan^3x)\sec^2x\cdot(\ldots)$, one reduces this (as in the source) to

$$I_2 = \int_0^{\pi/4}(\tan^3x-\tan^5x)\sec^2x\,dx = \left[\frac{\tan^4x}{4}-\frac{\tan^6x}{6}\right]_0^{\pi/4} = \frac1{12}$$

$7I_1+12I_2 = 0 + 1 = 1$

Solutions · 1