The integral _/6^/3^3x^23x(2^2x^23x+3 x6x)dx is equal to :

Question

The integral $\displaystyle\int_{\pi/6}^{\pi/3}	an^3x\cdot\sin^23x\left(2\sec^2x\cdot\sin^23x+3	an x\cdot\sin6xight)dx$ is equal to :

Accepted Answer

Recognise the integrand as the derivative of a product: note $\dfrac{d}{dx}(\tan^4x)=4\tan^3x\sec^2x$ and $\dfrac{d}{dx}(\sin^43x)=12\sin^33x\cos3x\cdot\ldots$, and the given expression matches $\dfrac12\dfrac{d}{dx}\left(\tan^4x\cdot\sin^43x\right)$ up to constants:
$$I=\frac12\int_{\pi/6}^{\pi/3}\frac{d}{dx}\left(\tan^4x\cdot\sin^43x\right)dx=\frac12\left[\tan^4x\sin^43x\right]_{\pi/6}^{\pi/3}$$

$$=\frac12\left(\tan^4\frac\pi3\cdot\sin^4\pi-\tan^4\frac\pi6\cdot\sin^4\frac\pi2\right)=\frac12\left(9\cdot0-\frac19\cdot1\right)=-\frac1{18}$$

Solutions · 1