Evaluate the following integrals: ( ^ -1 x) 1+x^2 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\sin(\tan^{-1}\text{x})}{1+\text{x}^2}\text{ dx}$

✓

Answer

$\int\frac{\sin(\tan^{-1}\text{x})}{1+\text{x}^2}\text{ dx}$ Let $\tan^{-1}\text{x}=\text{t}$ $\Rightarrow\frac{1}{1+\text{x}^2}\text{ dx}=\text{dt}$ Now, $\int\frac{\sin(\tan^{-1}\text{x})}{1+\text{x}^2}\text{ dx}$$=\int\sin\text{t dt}$
$=-\cos(\text{t})+\text{C}$
$=-\cos\big(\tan^{-1}\text{x}\big)+\text{C}$

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 Indefinite Integrals→Indefinite Integrals — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

Find the values of b for which the function $\text{f}(\text{x})=\sin\text{x}-\text{b}\text{x}+\text{c}$ is a decreasing function on R.

For each of the following p.d.f. of r.v. $X$, find (a) $P(X<1)$ and (b) $P(|X|<1)$ : $f(x)=\frac{x^2}{18}, \quad$ for $-3<x<3$ and $=0$, otherwise.

Find the general solutions of the following equations : cotθ = 0.

Diffrentiate the following w.r.t.x

$[\log \{\log (\log x)\}]^2$

Find a matrix X such that 2A + B + X = 0, where.
$\text{A}=\begin{bmatrix}-1&2\\3&4\end{bmatrix},\text{B}=\begin{bmatrix}3&-2\\1&5\end{bmatrix}$

Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this.
On R, define by $a^* b=a b^2$.
Here, $Z ^{+}$denotes the set of all non-negative integers.

Find the angle between the lines whose direction ratios are $4,-3,5$ and $3,4,5$.

Find the principal solutions of $\cos \theta=\frac{1}{2}$.

If $A$ and $B$ are symmetric matrices of the same order, write whether $AB − BA$ is symmetric or skew-symmetric or neither of the two.

Evaluate: $\int e^x\left[\frac{\sqrt{1-x^2} \sin ^{-1} x+1}{\sqrt{1-x^2}}\right] d x$