Question

integrate x ^ 5 * arcsin(x) dx from 0 to 1

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\int^{1}_{0}{x}^{5}\cdot\sin^{-1}(x)\,dx$

$\bf{Put\,\,\,x=sin(\theta)}$

$\bf{\mapsto\,\,dx=cos(\theta)\,d\theta}$

Now, limits become 0 ↔ π/2,

$\displaystyle\int^{\frac{\pi}{2}}_{0}{\sin}^{5}(\theta)\cdot\theta\cdot\cos(\theta)\,d\theta$

$\displaystyle=\int^{\frac{\pi}{2}}_{0}\theta\cdot{\sin}^{5}(\theta)\,\cos(\theta)\,d\theta$

$\displaystyle=\left[\theta\int{\sin}^{5}(\theta)\,\cos(\theta)\,d\theta\right]^{\frac{\pi}{2}}_{0}-\int^{\frac{\pi}{2}}_{0}\left[\dfrac{d}{d\theta}(\theta)\int{\sin}^{5}(\theta)\,\cos(\theta)\,d\theta\right]d\theta$

$\displaystyle=\left[\theta\cdot\dfrac{{\sin}^{6}(\theta)}{6}\right]^{\frac{\pi}{2}}_{0}-\int^{\frac{\pi}{2}}_{0}\left[1\cdot\dfrac{{\sin}^{6}(\theta)}{6}\right]d\theta$

$\displaystyle=\dfrac{\pi}{2}\cdot\dfrac{1}{6}-0-\dfrac{1}{6}\int^{\frac{\pi}{2}}_{0}{\sin}^{6}(\theta)\,d\theta$

$\displaystyle=\dfrac{\pi}{12}-\dfrac{1}{6}\int^{\frac{\pi}{2}}_{0}{\sin}^{6}(\theta)\,d\theta$

$\displaystyle=\dfrac{\pi}{12}-\dfrac{1}{6}\cdot\dfrac{5\cdot3\cdot1}{6\cdot4\cdot2}\times\dfrac{\pi}{2}$

$\displaystyle=\dfrac{\pi}{12}-\dfrac{1}{2}\cdot\dfrac{5\cdot1}{6\cdot4\cdot2}\times\dfrac{\pi}{2}$

$\displaystyle=\dfrac{\pi}{12}-\dfrac{5\,\pi}{192}$

$\displaystyle=\dfrac{16\,\pi-5\,\pi}{192}$

$\displaystyle=\dfrac{11\,\pi}{192}$