Evaluate the following integrals: ^ 2 _ - 2 ^3x dx - Vidyadip

Question

Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin^3\text{x dx}$

✓

Answer

Let $\text{I}=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin^3\text{x dx}$$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin\text{x}(1-\cos^2\text{x})\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin\text{x}\text{ dx}-\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin\text{x}\cos^2\text{x dx}$
$=\big[-\cos\text{x}\big]^{\frac{\pi}{2}}_{\frac{-\pi}{2}}+\Big[\frac{\cos^3\text{x}}{3}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$
$=0+0=0$

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 "Solve the Following Question.(2 Marks)" from Definite Integrals→Definite Integrals — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

Differentiate the following w. r. t. x.$\cos ^{-1}\left(\frac{1-9 x^2}{1+9 x^2}\right)$

Which one the following relations on A = {1, 2, 3} is a function?
f = {(1, 3), (2, 3), (3, 2)}, g = {(1, 2), (1, 3), (3, 1)}

Find the approximate values of:

$f(x)=x^3+5 x^2-7 x+10$ at $x=1.12$

$\int \frac{3 \cos x}{4 \sin ^2 x+4 \sin x-1} d x$

Evaluate the definite integral: $\int_0^{\frac{\pi}{2}} \cos ^2 x d x$

Evaluate : $\int_0^{\frac{\pi}{4}} \sec ^4 x d x$

If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the points $A, B, C$ respectively such that $3 \vec{a}+5 \vec{b}=8 \vec{c}$, then find the ratio in which $C$ divides $AB$.

If $A$ is a square matrix of order $3$ such that $adj\ (2A) = k\ adj\ (A), $ then write the value of $k.$

If $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$, then find the value of $x$

Question is modified

If $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{2}$, then find the value of $x$.

Write the domain of the real function $\text{f(x)}=\frac{1}{\sqrt{|\text{x}|-\text{x}}}.$