Question

evauluate the integral of int(cos²x sin³x)dx​

Answer 1

PLEASE REFER TO THE ATTACHMENT...

ACTUALLY THE INTEGRATION SIGN IS NOT AVAILABLE WITH ME

SO I DID IT IN THE COPY....

$\huge\star\mathtt\pink{HOPE\:IT\:HELPS}\star$

Answer 2

EXPLANATION.

⇒ ∫(cos²x sin³x)dx.

As we know that,

we can write equation as,

⇒ ∫(cos²x sin²x.sinx)dx.

As we know that,

Formula of,

⇒ sin²x = 1 - cos²x.

Put the value in equation, we get.

⇒ ∫(cos²x.(1 - cos²x).sin(x)).dx.

⇒ ∫(1 - cos²x)cos²x.sinx.dx.

Let we assume that,

⇒ cos(x) = t.

Differentiate w.r.t x, we get.

⇒ -sin(x)dx = dt.

Put the value in the equation, we get.

⇒ ∫(1 - t²)t²(-dt).

⇒ ∫(t² - t⁴)(-dt).

⇒ ∫(t⁴ - t²)dt.

⇒ ∫t⁴(dt) - ∫t²(dt).

⇒ t⁵/5 - t³/3 + c.

Substitute the value of t = cos(x) in equation, we get.

⇒ cos⁵(x)/5 - cos³(x)/3 + c.

                                                                                                                     

MORE INFORMATION.

(1) = ∫sin(x)dx = -cos(x) + c.

(2) = ∫cos(x)dx = sin(x) + c.

(3) = ∫tan(x)dx = ㏒(sec(x)) + c  Or  -㏒(cos(x)) + c.

(4) = ∫cot(x)dx = ㏒(sin(x)) + c.

(5) = ∫sec(x)dx = ㏒(sec(x) + tan(x)) + c  Or  -㏒(sec(x) - tan(x)) + c  Or  ㏒tan(π/4 + x/2) + c.

(6) = ∫cosec(x)dx = -㏒(cosec(x) + cot(x)) + c.  Or  ㏒(cosec(x) - cot(x)) + c  Or ㏒tan(x/2) + c.

(7) = ∫sec(x)tan(x)dx = sec(x) + c.

(8) = ∫cosec(x)cot(x)dx = -cosec(x) + c.

(9) = ∫sec²xdx = tan(x) + c.

(10) = ∫cosec²xdx = -cot(x) + c.