Try the fastest way to create flashcards
Question

Evaluate using Integration by Parts as a first step.

sin1xx2dx\int \frac { \sin ^ { - 1 } x } { x ^ { 2 } } d x

Solution

Verified
Step 1
1 of 6

We need to evaluate the integral :

sin1xx2 dx\int \dfrac{\sin^{-1} x}{x^2}\ dx

We use method of integration by parts for solving this integral

u dv=uvv du\begin{gather} \int u\ dv=uv-\int v\ du \end{gather}

Here, we take u=sin1xu =\sin^{-1} x and dv=1x2 dxdv= \dfrac{1}{x^2}\ dx. So v=x2+12+1=1xv=\dfrac{x^{-2+1}}{-2+1} = -\dfrac 1x and

ddxsin1x=11x2    du=11x2 dx\dfrac{d}{dx}\sin^{-1} x =\dfrac{1}{\sqrt{1-x^2}} \implies du = \dfrac{1}{\sqrt{1-x^2}}\ dx

Create a free account to view solutions

Create a free account to view solutions

Recommended textbook solutions

Thomas' Calculus 14th Edition by Christopher E Heil, Joel R. Hass, Maurice D. Weir

Thomas' Calculus

14th EditionISBN: 9780134438986 (11 more)Christopher E Heil, Joel R. Hass, Maurice D. Weir
10,142 solutions
Calculus: Early Transcendentals 8th Edition by James Stewart

Calculus: Early Transcendentals

8th EditionISBN: 9781285741550 (6 more)James Stewart
11,081 solutions
Calculus: Early Transcendentals 4th Edition by Colin Adams, Jon Rogawski, Robert Franzosa

Calculus: Early Transcendentals

4th EditionISBN: 9781319050740 (2 more)Colin Adams, Jon Rogawski, Robert Franzosa
8,977 solutions
Calculus: Early Transcendentals 9th Edition by Daniel K. Clegg, James Stewart, Saleem Watson

Calculus: Early Transcendentals

9th EditionISBN: 9781337613927 (3 more)Daniel K. Clegg, James Stewart, Saleem Watson
11,050 solutions

More related questions

1/4

1/7