How do you integrate #tan^3xsec^4xdx#?

Question

Aurora Alvarado · Accepted Answer

#inttan³xsec⁴xdx=1/(6cos^6(x))-1/(4cos^4(x))+C#, #C in RR# #inttan³xsec⁴xdx# #=int(sin³(x))/(cos^7(x))dx# #=-int(-sin(x)(1-cos²(x)))/(cos^7(x))dx# Let #u=cos(x)# #du=-sin(x)dx# #<=> -int(1-u²)/(u^7)du# #=-int1/u^7du+int1/u^5du# #=1/(6u^6)-1/(4u⁴)# #=1/(6cos^6(x))-1/(4cos^4(x))+C#, #C in RR# \0/ here's our answer !

Axel Alvarez · Answer

undefined To integrate $ \tan^3(x) \sec^4(x) \, dx $, you can use integration by parts combined with trigonometric identities.

Let:
$$ u = \tan(x) \quad \text{and} \quad dv = \sec^4(x) \, dx $$

Then, compute $ du $ and $ v $:
$$ du = \sec^2(x) \, dx \quad \text{and} \quad v = \frac{\tan(x) \sec^2(x)}{3} + \frac{2 \tan^3(x) \sec^2(x)}{3} $$

Now, apply integration by parts formula:
$$ \int u \, dv = uv - \int v \, du $$

Substitute the values:
$$ \int \tan^3(x) \sec^4(x) \, dx = \frac{\tan(x) \sec^3(x)}{3} + \frac{2 \tan^3(x) \sec^2(x)}{3} - \frac{1}{3} \int \tan(x) \sec^2(x) \, dx $$

The last integral on the right-hand side, $ \int \tan(x) \sec^2(x) \, dx $, can be evaluated using a substitution $ u = \sec(x) $ or by recognizing that its derivative is $ \sec(x) \tan(x) $.

$$ \int \tan^3(x) \sec^4(x) \, dx = \frac{\tan(x) \sec^3(x)}{3} + \frac{2 \tan^3(x) \sec^2(x)}{3} - \frac{1}{3} \left( \frac{\sec(x) \tan(x)}{2} - \frac{\ln|\sec(x) + \tan(x)|}{2} \right) + C $$

Where $ C $ is the constant of integration.