Question
Evaluate the following integrals:$\int\limits^{\pi}_0\text{x}\sin^3\text{x}\text{ dx}$
✓
Answer
Let $\text{I}=\int\limits^{\pi}_0\text{x}\sin^3\text{x}\text{ dx}\ ...(\text{i})$$=\int\limits^{\pi}_0(\pi-\text{x})\sin^3(\pi-\text{x})\text{dx}$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin^3\text{x}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\sin^3\text{x}\text{ dx}$
$=\int\limits^{\pi}_0\pi\sin^3\text{x}\text{ dx}$
$=\int\limits^{\pi}_0\pi\frac{3\sin\text{x}-\sin3\text{x}}{4}\text{ dx}$
$=\frac{\pi}{4}\int\limits^{\pi}_0\big(3\sin\text{x}-\sin3\text{x}\big)\text{dx}$
$=\frac{\pi}{4}\Big[-3\cos\text{x}+\frac{\cos3\text{x}}{3}\Big]^{\pi}_0$
$=\frac{\pi}{4}\Big[-3\cos\pi+3\cos0+\frac{\cos3\pi}{3}-\frac{\cos0}{3}\Big]$
$=\frac{\pi}{4}\Big[3+3+\frac{-1}{3}-\frac{1}{3}\Big]$
$=\frac{\pi}{4}\Big[3-\frac{1}{3}\Big]$
$=\frac{\pi}{2}\times\frac{8}{3}$
$=\frac{4\pi}{3}$
$\therefore\ \text{I}=\frac{2\pi}{3}$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin^3\text{x}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\sin^3\text{x}\text{ dx}$
$=\int\limits^{\pi}_0\pi\sin^3\text{x}\text{ dx}$
$=\int\limits^{\pi}_0\pi\frac{3\sin\text{x}-\sin3\text{x}}{4}\text{ dx}$
$=\frac{\pi}{4}\int\limits^{\pi}_0\big(3\sin\text{x}-\sin3\text{x}\big)\text{dx}$
$=\frac{\pi}{4}\Big[-3\cos\text{x}+\frac{\cos3\text{x}}{3}\Big]^{\pi}_0$
$=\frac{\pi}{4}\Big[-3\cos\pi+3\cos0+\frac{\cos3\pi}{3}-\frac{\cos0}{3}\Big]$
$=\frac{\pi}{4}\Big[3+3+\frac{-1}{3}-\frac{1}{3}\Big]$
$=\frac{\pi}{4}\Big[3-\frac{1}{3}\Big]$
$=\frac{\pi}{2}\times\frac{8}{3}$
$=\frac{4\pi}{3}$
$\therefore\ \text{I}=\frac{2\pi}{3}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
A box manufacturer makes large and small boxes from a large piece of cardboard. The large boxes require 4 sq. metre per box while the small boxes require 3 sq. metre per box. The manufacturer is required to make at least three large boxes and at least twice as many small boxes as large boxes. If 60 sq. metre of cardboard is in stock, and if the profits on the large and small boxes are Rs. 3 and Rs. 2 per box, how many of each should be made in order to maximize the total profit?
If $\text{y}=\big\{\log_{\cos\text{x}}\sin\text{x}\big\}\big\{\log_{\sin\text{x}}\cos\text{x}\big\}^{-1}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big),$ find $\frac{\text{dy}}{\text{dx}}$ at $\text{x}=\frac{\pi}{4}$
→Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$4\text{x}+3\text{y}=\log\big(4\text{x}-3\text{y}\big)$
→$4\text{x}+3\text{y}=\log\big(4\text{x}-3\text{y}\big)$
Evaluate the following integrals:
$\int\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}\text{dx}$
→$\int\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}\text{dx}$
Find tha area bounded by the curves $x = y^2$ and $x = 3 - 2y^2.$
→Write the following in the simplest form:
$\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-1}{\text{x}}\Big\},\text{x}\neq0$
→$\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-1}{\text{x}}\Big\},\text{x}\neq0$
We first draw the lines processed on three machines : Assembling, Finishing and Polishing. The time required for each product in hours and availability of each machine is given by following table:
Formulate the above problem as L.P.P. Solve it graphically to get maximum profit.
Find the inverse of the following matrices by using elementry row transformation:$\begin{bmatrix}3 & 10 \\ 2 & 7 \end{bmatrix}$
→Find the maximum and the minimum values, if any, without using derivaives of the following functions:$f(x) = |x + 2| + 2$ on $R$.
→If the median to the base of a triangle is perpendicular to the base, then triangle is isosceles.