Question

Need help finding following integrals!
fxsin^-1xdx

f(pi/4-9) tan^5xsec^3xdx

f(3-2) (x^3+1)/(x^3-x^2)dx

Answer 1

Anyone?

Answer 2

okay, you'll have to integrate by parts.

Answer 3

can you show me step by step

Answer 4

if the first one is , ∫ x / sin x dx, then i have no idea.

if it is ∫ x arcsin x dx, then you could do worse than make a substitution: u = arcsinx, x = sin u, dx = cosu du.

then a double angle formula to combine the sin and cos terms into a sin.

then a straightforward integration by parts.

Answer 5

your numbers show up as question marks

Answer 6

refresh

Answer 7

hmm

Answer 8

i need to see the step by step

Answer 9

You can also approach \(\int x\sin^{-1}x\,dx\) immediately using IBP, with
\[\begin{matrix}u=\sin^{-1}x&&&dv=x\,dx\\du=\dfrac{dx}{\sqrt{1-x^2}}&&&v=\dfrac{1}{2}x^2\end{matrix}\]
Then
\[\int x\sin^{-1}x\,dx=\frac{1}{2}x^2\sin^{-1}x-\frac{1}{2}\int\frac{x^2}{\sqrt{1-x^2}}\,dx\]
essentially the reverse of what IrishBoy suggested.

For the second integral, notice that setting \(y=\ln (\sin x)\) gives \(dy=\dfrac{\cos x}{\sin x}\,dx=\cot x\,dx\).

For the third, use your identities:
\[\tan^5x\sec^3x=(\sec^2x-1)^2\sec^2x\color{red}{\sec x\tan x}\]
The red factor is particularly useful, since setting \(t=\sec x\) yields \(dt=\sec x\tan x\,dx\), and so
\[\int_0^{\pi/4} \tan^5x\sec^3x\,dx=\int_1^{\sqrt2} t^2(t^2-1)^2\,dt\]
which can be handled in a variety of ways. Expanding might be the most obvious route, but a substitution might also be of some help, whether it be algebraic or trigonometric.

For the fourth integral, consider using long division to "simplify" the integrand, then break up what you can into partial fractions.