Question

integral of sin^-1(2x)​

Answer 1

Answer:

$\large\boxed{\sf{I=x \: { \sin }^{ - 1} (2x) + \frac{ \sqrt{1 - 4 {x}^{2} } }{2} + c}}$

Step-by-step explanation:

$I=\displaystyle \int { \sin }^{ - 1} (2x)dx$

Let's assune that

$2x = \sin( \theta)$

$= > 2dx = \cos( \theta) d \theta$

Therefore, we get

$I=\displaystyle \int \frac{1}{2} { \sin }^{ - 1} ( \sin \theta) \cos \theta d \theta \\ \\ = \frac{1}{2} \displaystyle \int \theta \cos \theta d \theta$

It's of the form $\displaystyle \int (uv) dx$ .

Solving further, we get

Refer to the attachment

Hence , we get,

$I= x \: { \sin }^{ - 1} (2x) + \frac{ \sqrt{1 - 4 {x}^{2} } }{2} + c$

Answer 2

Answer:

Step-by-step explanation:

I=\displaystyle \int { \sin }^{ - 1} (2x)dxI=∫sin

−1

(2x)dx

Let's assune that

2x = \sin( \theta)2x=sin(θ)

= > 2dx = \cos( \theta) d \theta=>2dx=cos(θ)dθ

Therefore, we get

\begin{lgathered}I=\displaystyle \int \frac{1}{2} { \sin }^{ - 1} ( \sin \theta) \cos \theta d \theta \\ \\ = \frac{1}{2} \displaystyle \int \theta \cos \theta d \theta\end{lgathered}

I=∫

2

1

sin

−1

(sinθ)cosθdθ

=

2

1

∫θcosθdθ

It's of the form \displaystyle \int (uv) dx∫(uv)dx .

Solving further, we get

Refer to the attachment

Hence , we get,

I= x \: { \sin }^{ - 1} (2x) + \frac{ \sqrt{1 - 4 {x}^{2} } }{2} + cI=xsin

−1

(2x)+

2

1−4x

2

+c

follow