What is the integral of tan^4(x)cos^3(x) ?

The integral of tan^4(x)cos^3(x) is -(sin^3(x))/3-sin(x)+(ln|sin(x)+1|)/2-(ln|sin(x)-1|)/2+C

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integral of tan^4(x)cos^3(x)

\int tan^{4} (x) cos^{3} (x) d x

- \frac{sin ^{3} ( x )}{3} - sin (x) + \frac{ln ∣ sin ( x ) + 1 ∣}{2} - \frac{ln ∣ sin ( x ) - 1 ∣}{2} + C

Solution steps

\int tan^{4} (x) cos^{3} (x) d x

Rewrite using trig identities

= \int \frac{sin ^{4} ( x )}{cos ( x )} d x

Apply Trigonometric Substitution

: \int \frac{u ^{4}}{1 - u ^{2}} d u

= \int \frac{u ^{4}}{1 - u ^{2}} d u

Long division

\frac{u ^{4}}{1 - u ^{2}} : - u^{2} - 1 + \frac{1}{- u ^{2} + 1}

= \int - u^{2} - 1 + \frac{1}{- u ^{2} + 1} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int u^{2} d u - \int 1 d u + \int \frac{1}{- u ^{2} + 1} d u

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= - \frac{u ^{3}}{3} - u + \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

Substitute back

u = sin (x)

= - \frac{sin ^{3} ( x )}{3} - sin (x) + \frac{ln ∣ sin ( x ) + 1 ∣}{2} - \frac{ln ∣ sin ( x ) - 1 ∣}{2}

Add a constant to the solution

= - \frac{sin ^{3} ( x )}{3} - sin (x) + \frac{ln ∣ sin ( x ) + 1 ∣}{2} - \frac{ln ∣ sin ( x ) - 1 ∣}{2} + C

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What is the integral of tan^4(x)cos^3(x) ?
The integral of tan^4(x)cos^3(x) is -(sin^3(x))/3-sin(x)+(ln|sin(x)+1|)/2-(ln|sin(x)-1|)/2+C