How do I solve this trig integral?

Question

I don't know how to solve the integral of Cos^2(x) Tan^3(x) dx. Can someone show me how?

Thanks...

I don't know how to solve the integral of Cos^2(x) Tan^3(x) dx. Can someone show me how?

Thanks...

So... do I let U=Sin^2(x) and dV=Tan(x)? And will dU be 1/2(Sin(x))Cos(x) using chain rule?

So... do I let U=Sin^2(x) and dV=Tan(x)? And will dU be 1/2(Sin(x))Cos(x) using chain rule?

New to Yahoo Answers. Sorry hahah

Answer 1

∫ (cos²(x) tan³(x) dx)

To solve this, we must first manipulate it into a better integrable form. tan³(x) = sin³(x)/cos³(x), so

cos²(x) tan³(x) = cos²(x) ( sin³(x)/cos³(x) )

This reduces to

= sin³(x)/cos(x)

So let's integrate this.

∫ ( sin³(x)/cos(x) dx)

To solve this integral, first, I'm going to break off a sin(x) from the sin³(x). I'm going to move sin(x) next to the dx.

∫ ( sin²(x)/cos(x) sin(x) dx)

Use the identity sin²(x) = 1 - cos²(x).

∫ ( { [ 1 - cos²(x) ] / cos(x) } sin(x) dx )

Here's where we use u-substitution.
Let u = cos(x). Then
du = -sin(x) dx, and
(-1) du = sin(x) dx.
{Note: The tail end of our integral is sin(x) dx, so it follows that the tail end after our substitution will be (-1) du.}

∫ ( { [ 1 - u² ] / u } (-1) du )

Normally I would pull the constant (-1) out, but instead, I'm going to apply it to the term in brackets and therefore flip them. After all -(a - b) is the same as (b - a).

∫ ( { [ u² - 1 ] / u } du )

We can easily solve this by splitting this up into two fractions and reducing.

∫ ( u²/u - 1/u ) du

∫ (u - 1/u) du

And now, this is easily integrable.

(1/2)u² - ln|u| + C

But u = cos(x), so our final answer is

(1/2)cos²(x) - ln|cos(x)| + C

Answer 2

Cos^2(x) Tan^3(x) dx = Cos^2(x) [sin^3(x)/cos^3(x)] dx = sin^3(x)/cos(x)dx = sin^2(x)tan(x)dx

Now integrate by parts taking into acount that the integral of tan(x) = -log(cos(x). This integration has to be performes twice.

The result is - sin^(x)/2 - ln[cos(x)]

Answer 3

cos²(x) tan³(x) =
sin³(x) / cos(x) =
sin(x) (1-cos²(x))/cos(x) =
tan(x) - sin(x)cos(x)

So ∫cos²(x) tan³(x) dx =
-ln(cos(x)) - sin²(x)/2 + C

Answer 4

ok.. i rather loved Mercedes' strategies-set yet think of the final term must be sin(4x) no longer cos(4x). She did get the main superb answer. (cos)^4=((cosx)^2)^2= ((a million+cos2x)/2]^2 (a million/4) [a million+2cos(2x)+(cos(2x))^2] (a million/4) [a million+2cos(2x)+(a million+cos(4x))/2] (a million/4) [ x + sin(2x) +x/2+ SIN(4x)/8] x/4+sin(2x)/4+x/8+SIN(4x)/32 from 0 to pi/2 3x/8+sin(2x)/4+SIN(4x)/32 from 0 to pi/2 = 3pi/sixteen +0 +0 - 0- 0 - 0. = 3pi/sixteen