Question

Integral((sinx)^4*(COSX)^-4dx

Answer 1

$\int \sin^{4} (x) \times \cos^{ - 4} (x) dx = \int \sin^{4} (x) \times \frac{1}{ \cos^{4} (x) } dx =$

$\int \frac{ \sin^{4}(x ) }{ \cos^{4} (x) } dx = \int \tan^{4} (x) dx$

По формуле:

$\int \tan^{n} (x) dx = \frac{1}{n - 1} \times \tan^{n - 1} (x) - \int \tan^{n - 2} (x)$

$\frac{1}{4 - 1} \times \tan^{4 - 1} (x) - \int \tan^{4 - 2} (x) = \frac{1}{3} \times \tan^{3} (x) - \int \tan^{2} (x)$

Раскроем tan²x по формул

$\tan^{2} (x) = \frac{1}{ \cos^{2} (x) } - 1$

$\frac{1}{3} \times \tan^{3} (x) - \int \frac{1}{ \cos^{2} (x) } - 1dx =$

$\frac{1}{3} \times \tan^{3} (x) - (\tan(x) - x) = \frac{ \tan ^{3} (x) }{3} - \tan(x) + x$

И конечный ответ:

$\int \sin^{4} (x) \times \cos^{ - 4} (x) dx = \frac{ \tan ^{3} (x) }{3} - \tan(x) + x + c$