Question #29a24

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Answer 1 · 2016-12-10T17:50:53

$\int 6 {sin}^{2} (x) {cos}^{- 4} (x) d x = 2 {tan}^{3} (x) + C$ $int6sin^2(x)cos^-4(x)dx = 2tan^3(x) + C$

$\int 6 {sin}^{2} (x) {cos}^{- 4} (x) d x$ $int6sin^2(x)cos^-4(x)dx$

Use the identities $tan (x) = \frac{sin (x)}{cos (x)} and sec (x) = \frac{1}{cos (x)}$ $tan(x) = sin(x)/cos(x) and sec(x) = 1/cos(x)$ :

#6inttan^2(x)sec^2(x)dx

Let #u = tan(x)", then, "du = sec^2(x)dx

Then the integral becomes:

$6 \int u^{2} d u$ $6intu^2du$

This is a trivial integration:

$\frac{6}{3} u^{3} + C$ $6/3u^3 + C$

Reverse the substitution:

$2 {tan}^{3} (x) + C$ $2tan^3(x) + C$

The answer is:

$\int 6 {sin}^{2} (x) {cos}^{- 4} (x) d x = 2 {tan}^{3} (x) + C$ $int6sin^2(x)cos^-4(x)dx = 2tan^3(x) + C$