Download App

INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
Evaluate the following integrals:
$\int^\limits{\pi}_0\sin^3\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
✓

Answer

Let $\text{I}=\int^\limits{\pi}_0\sin^3\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$ Then,
$\text{I}=\int^\limits{\pi}_0\sin\text{x }\sin^2\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
$\Rightarrow\text{I}=\int^\limits{\pi}_0\sin\text{x}(1-2\cos^2\text{x})(1+\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
$\Rightarrow\text{I}=\int^\limits{\pi}_0\sin\text{x}(1-\cos\text{x})(1+2\cos\text{x})(1+\cos\text{x})^3\text{ dx}$
Let $\cos\text{x}=\text{t}$ Then, $-\sin\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=1$ and $\text{x}=\pi,\text{t}=-1$
$\therefore\ \text{I}=-\int^\limits{-1}_1(1-\text{t})(1+2\text{t})(1+\text{t})^3\text{ dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+\text{t}-2\text{t}^2\big)\big(1+\text{t}^3+3\text{t}+3\text{t}^2\big)\text{dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+\text{t}^3+3\text{t}+3\text{t}^3+\text{t}+\text{t}^4\\+3\text{t}^2+3\text{t}^3-2\text{t}^2-2\text{t}^5-6\text{t}^3-6\text{t}^4\big)\text{dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+4\text{t}+4\text{t}^2-2\text{t}^3-5\text{t}^4-2\text{t}^5\big)\text{dt}$
$\Rightarrow\text{I}=\Big[\text{t}+2\text{t}^2+\frac{4\text{t}^3}{3}-\frac{\text{t}^4}{2}-\text{t}^5-\frac{\text{t}^6}{3}\Big]^1_{-1}$
$\Rightarrow\text{I}=1+2+\frac{4}{3}-\frac{1}{2}-1-\frac{1}{3}\\+1-2+\frac{4}{3}+\frac{1}{2}-1+\frac{1}{3}$
$\Rightarrow\text{I}=\frac{8}{3}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from INTEGRALS→INTEGRALS — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Evaluate the following intregals:
$\int\frac{\text{x}^2+\text{x}-1}{(\text{x}+1)^2(\text{x}+2)}\ \text{dx}$
→
$\text{Evaluate:}\int\frac{\text{x + 2}}{\sqrt{\text{x}^{2}+\text{2x}+}\text{3}}\text{dx}$
→
Show that the lines $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and $\frac{x-2}{1}=\frac{y+4}{3}=\frac{z-6}{5}$ intersect each other. Find also the coordinates of the point of intersection.
→
Evalute the following integrals:
$\int\frac{1}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$
→
Find the equation of the tangent line to the curve $y = x^2 - 2x + 7$ which is perpendicular to the line $5y - 15x = 13.$
→
Integrate the function $ \frac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}$
→
Find the equation of the curve satisfying $\text{x}(\text{x}+1)\frac{\text{dy}}{\text{dx}}-\text{y}=\text{x}(\text{x}+1)$ and passing through (1, 0).
→
Find the relationship between 'a' and 'b' so that the function 'f' defind by $\text{f(x)}=\begin{cases}\text{ax}+1,&\text{if }\text{ x}\leq3\\\text{bx}+3,&\text{if }\text{ x}>3\end{cases}$ is continuous at x = 3.
→
Let $\text{A}=\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}$ and $\text{B}=\begin{bmatrix}0&4&3\\1&-3&-3\\-1&4&4\end{bmatrix},$ compute $A^2 - B^2$.
→
Evaluate the following integrals:$\int\frac{1}{\cos\text{x}+\text{cosec x}}\text{dx}$
→