How do you find the indefinite integral of ∫ dx / sin^3 x cos^5 x ?

Question

How do you find the indefinite integral of ∫ dx / sin^3 x cos^5 x ?

1 Answer

Impact of this question

9739 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-15T09:16:37

$I = \frac{1}{4} {tan}^{4} (x) + \frac{3}{2} {tan}^{2} (x) + 3 ln (tan (x)) - \frac{1}{2} {cot}^{2} (x) + C$ $I=1/4tan^4(x)+3/2tan^2(x)+3ln(tan(x))-1/2cot^2(x)+C$

Explanation:

We want to solve

$I = \int \frac{1}{{sin}^{3} (x) {cos}^{5} (x)} d x$ $I=int1/(sin^3(x)cos^5(x))dx$

Multiply the DEN and NUM by ${sec}^{8} (x)$ $sec^8(x)$ ,
an integral consisting of tangens and secant have good opputunities substitution

$I = \int \frac{{sec}^{8} (x)}{{tan}^{3} (x)} d x$ $I=intsec^8(x)/(tan^3(x))dx$

Make a substitution $u = tan (x) \Rightarrow d u = {sec}^{2} (x) d x$ $color(red)(u=tan(x)=>du=sec^2(x)dx$

$I = \int \frac{{sec}^{6} (x)}{u^{3}} d u$ $I=intsec^6(x)/(u^3)du$

Use the identity ${sec}^{2} (x) = {tan}^{2} (x) + 1$ $color(blue)(sec^2(x)=tan^2(x)+1$

$I = \int \frac{{(u^{2} + 1)}^{3}}{u^{3}} d u$ $I=int(u^2+1)^3/(u^3)du$

$I = \int \frac{u^{6} + 3 u^{4} + 3 u^{2} + 1}{u^{3}} d u$ $color(white)(I)=int(u^6+3u^4+3u^2+1)/(u^3)du$

$I = \int u^{3} + 3 u + \frac{3}{u} + \frac{1}{u^{3}} d u$ $color(white)(I)=intu^3+3u+3/u+1/u^3du$

$I = \frac{1}{4} u^{4} + \frac{3}{2} u^{2} + 3 ln (u) - \frac{1}{2 u^{2}} + C$ $color(white)(I)=1/4u^4+3/2u^2+3ln(u)-1/(2u^2)+C$

Substitute back $u = tan (x)$ $u=tan(x)$

$I = \frac{1}{4} {tan}^{4} (x) + \frac{3}{2} {tan}^{2} (x) + 3 ln (tan (x)) - \frac{1}{2} {cot}^{2} (x) + C$ $I=1/4tan^4(x)+3/2tan^2(x)+3ln(tan(x))-1/2cot^2(x)+C$

Science

Math

Humanities

... and beyond

How do you find the indefinite integral of ∫ dx / sin^3 x cos^5 x ?

1 Answer

Explanation:

Related questions

Impact of this question