How do you integrate these 2 questions. They involve the same steps, one is with trig and one is without. I'm curious how the 2 equations differ. thanks!  `int(3x^3)csc(pix^4)cot(pix^4)dx`...

How do you integrate these 2 questions. They involve the same steps, one is with trig and one is without. I'm curious how the 2 equations differ. thanks! 

`int(3x^3)csc(pix^4)cot(pix^4)dx`

`int(5x-1)/sqrt(3-4x^2)dx`
` `

Asked on by moocow554

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lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

(#2) `int(5x-1)/sqrt(3-4x^2)dx`

Express it as difference of two integrals.

`=int(5x)/sqrt(3-4x^2) -int1/sqrt(3-4x^2)dx`

Since there is no x at the numerator of the second integral, express its denominator as difference of two squares.

`=int(5x)/sqrt(3-4x^2)-int1/sqrt((sqrt3)^2-(2x)^2)dx`

Use u-substitution for each integral.

For the first integral, it is:

`u=3-4x^2`

`du=-8xdx`

`-(du)/8=xdx`

For the second integral, it is:

`u=2x`

` ` `du=2dx`

`(du)/2=dx`

Then, plug-in them to the two integrals.

`=int 5/sqrtu (-(du)/8) -int1/sqrt((sqrt3)^2-u^2)((du)/2)`

`=-5/8int1/sqrtu du - 1/2int (du)/sqrt((sqrt3)^2-u^2)`

`=-5/8int u^(-1/2)du -1/2int (du)/sqrt((sqrt3)^2-u^2)`

` `For the first integral apply the power formula `int u^ndu =u^(n+1)/(n+1)+C`  .

For the second integral, apply the formula `int (du)/sqrt(a^2-u^2)=sin^(-1)(u/a) + C`.

`=-5/8*u^(1/2)/(1/2)-1/2sin^(-1)(u/sqrt3) +C`

`=-5/8*2/1sqrtu-1/2sin^(-1)(u/sqrt3)+C`

`=-5/4sqrtu -1/2sin^(-1)(u/sqrt3)+C`

And, substitute back the u's of the firt and second integral.

`=-5/4sqrt(3-4x^2)-1/2sin^(-1)((2x)/sqrt3)+C` Therefore, `int (5x-1)/sqrt(3-4x^2)dx=-5/4sqrt(3-4x^2) -1/2sin^(-1)((2x)/sqrt3)+C` .

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lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

(#1)`int(3x^3)csc(pix^4)cot(pix^4)dx`

Since the integrand has a form csc(u)cot(u), consider the integral formula

`intcsc(u)cot(u)du=-cscu+C`

To be able to apply that, use u-substitution method.

Let,

`u=pix^4`

Then, differentiate it.

`du=4pix^3dx`

Since there is a factor x^3 in the integrand, isolate x^3dx.

`(du)/(4pi)=x^3dx`

Then, plug-in   `u=pix^4`   and  `(du)/(4pi)=x^3dx`   to the integrand above.

`int(3x^3)csc(pix^4)cot(pix^4)dx`

`=int3csc(u)cot(u)((du)/(4pi))`

`=3/(4pi)intcsc(u)cot(u)du`

`=3/(4pi)(-csc(u))+C`

`=-3/(4pi)cscu+C`

And, substitute back  `u=pix^4` .

`=-3/(4pi)csc(pix^4)+C`

Therefore, `int(3x^3)csc(pix^4)cot(pix^4)dx=-3/(4pi)csc(pix^4)+C`

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