I have these two integrals that I need to solve by using trig substitution, I di
ID: 2828896 • Letter: I
Question
I have these two integrals that I need to solve by using trig substitution, I did the substitution and I'm stuck in the step that comes after that. I know there may be alternative ways to do it but I need to solve them by using the trig substitutions...
a) Integral of [(1-x^2)^(3/2)dx] / (x^6 )
I let x=sin? so dx= cos? d?
After performing the trig substitution and applying a Pythagorean identity I end up with Integral of (cos^4? d?) / (sin^6?), how do I solve from here?
b) Integral of (x^2 dx) / [(1-x^2)^(5/2)]
I let x=sin ? so dx = cos? d?
After performing the trig substitution and applying a Pythagorean identity I end up with Integral of (sin^2? d?) / (cos^4?), how do I solve from here?
Thank you!!!!!!
Explanation / Answer
a) Integral of [(1-x^2)^(3/2)dx] / (x^6 )
x = sin (theta) => dx = cos ( theta ) d (theta )
=>integral [ cos^3 (theta ) ] cos ( theta ) d (theta ) / sin^6 (theta)
= integral [ cos^4 (theta) ] / sin^6 (theta)
= integral [ cos (theta) / sin (theta) ] ^4 csc^2 (theta) d (theta)
= integral [ cot^4 (theta) ] csc^2 (theta) d ( theta )
let cot (theta) = t => -csc^2 (theta) d(theta) = dt
= integral [ - t^4 dt ]
= -t^5 / 5 + c
= - [ cot (theta) ]^5 / 5 + c
= - [ cot ( sin^-1 (x) ] ^5 / 5 + c
= - [ cot [ cot^-1 (sqrt(1-x^2 ) / x) ]^5 / 5 + c
= - [( sqrt(1-x^2 )) / x )^5 / 5 + c
= - [( 1-x^2)^(5/2) ] / 5x^5 + c
b) Integral of (x^2 dx) / [(1-x^2)^(5/2)]
x = sin ( theta ) => dx = cos ( theta ) d ( theta)
= integral [ sin^2 ( theta ) cos ( theta ) d ( theta ) ] / [ cos^5 ( theta ) ]
= integral [ sin^2 ( theta ) / cos^4 ( theta ) ] d ( theta )
= integral [ tan^2 ( theta ) sec^2 ( theta ) ] d ( theta )
tan ( theta ) = t => sec^2 ( theta ) d ( theta ) = dt
= integral [ t^2 dt ]
= t^3 /3 + c
= ( tan ( theta ) )^3 / 3 + c
= [ tan ( sin^-1 ( x ) ] ^3 / 3 + c
= [ tan ( tan^-1 x / sqrt(1-x^2 ) ] ^3 / 3 + c
= [ x / sqrt(1-x^2) ] ^3 /3 +c
= [ x^3 ] / 3[ 1 - x^2 ]^(3/2) + c