What is the integral of tan squared x cosine to the 5th power x dx?

Answer 1

Since you did not specify any limits of integration, I assume you are looking for the indefinite integral of this expression:

tan2(x)cos5(x)

with respect to x (dx). Using the following identity:

tan(x) = sin(x) / cos(x)

The original expression can be rewritten as:

(sin2(x) / cos2(x))cos5(x)

Which further simplifies to:

sin2(x)cos3(x)

Which can be expanded to:

sin2(x)cos2(x)cos(x)

Using the identity:

sin2(x) + cos2(x) = 1

which implies:

cos2(x) = 1 - sin2(x)

which makes the expression from above able to be simplified into:

sin2(x)(1 - sin2(x))cos(x)

From here, you can use u-substitution by using the substitution:

u = sin(x)

du = cos(x) dx => dx = du/cos(x)

So after u substitution:

int(sin2(x)(1 - sin2(x))cos(x)) dx

becomes:

int(u2(1-u2)) du

int(u2-u4) du

From here, elementary antiderivatives can be used:

anti(u2) = (1/3)(u3)

anti(u4) = (1/5)(u5)

which yields a final indefinite integral in u of:

(1/3)u3-(1/5)u5 + C

where C is the constant of integration (since this is an indefinite integral).

Back-substituting with the u-substitution from before (u=sin(x)), the final indefinite integral in x is:

(1/3)sin3(x) - (1/5)sin5(x) + C