# Thread: Integrate cos^2 t using integration by parts

1. ## Integrate cos^2 t using integration by parts

How do we do this, please? Thanks.

Find $\displaystyle \int cos(t)^2\ dt$.

Let $\displaystyle u=cos(t), du=-sin(t)dt, v=sin(t), dv=cos(t)dt$.

$\displaystyle \int cos^2(t) dt \\=\int u\ dv \\= uv-\int v\ du \\= cos(t)sin(t)-\int -sin(t)^2\ dt \\= cos(t)sin(t)+\int sin(t)^2\ dt.$

Let $\displaystyle u=sin(t), du=cos(t)dt, v=-cos(t), dv=sin(t)dt$.

$\displaystyle cos(t)sin(t)+\int sin(t)^2\ dt \\=cos(t)sin(t)+\int u\ dv \\=cos(t)sin(t) + \left[uv-\int v\ du \right] \\=cos(t)sin(t) + \left[-cos(t)sin(t) + \int cos(t)^2\ dt \right] \\=\int cos(t)^2\ dt.$

Thus proving, once and for all, that
$\displaystyle \int cos(t)^2\ dt=\int cos(t)^2\ dt$

2. ## Re: Integrate cos^2 t using integration by parts

You can say $\displaystyle \displaystyle \sin^2 t = \frac{1-\cos 2t}{2}$ might make it easier.

3. ## Re: Integrate cos^2 t using integration by parts

How do we do this, please? Thanks.

Find $\displaystyle \int cos(t)^2\ dt$.

Let $\displaystyle u=cos(t), du=-sin(t)dt, v=sin(t), dv=cos(t)dt$.

$\displaystyle \int cos^2(t) dt \\=\int u\ dv \\= uv-\int v\ du \\= cos(t)sin(t)-\int -sin(t)^2\ dt \\= cos(t)sin(t)+\int sin(t)^2\ dt.$

Let $\displaystyle u=sin(t), du=cos(t)dt, v=-cos(t), dv=sin(t)dt$.

$\displaystyle cos(t)sin(t)+\int sin(t)^2\ dt \\=cos(t)sin(t)+\int u\ dv \\=cos(t)sin(t) + \left[uv-\int v\ du \right] \\=cos(t)sin(t) + \left[-cos(t)sin(t) + \int cos(t)^2\ dt \right] \\=\int cos(t)^2\ dt.$

Thus proving, once and for all, that
$\displaystyle \int cos(t)^2\ dt=\int cos(t)^2\ dt$
If you MUST use integration by parts (which is the most tedious method, when, as Pickslides says, the double angle formula for cosine simplifies the integrand greatly)...

Let \displaystyle \displaystyle \begin{align*} u = \cos{x} \implies du = -\sin{x}\,dx \end{align*} and \displaystyle \displaystyle \begin{align*} dv = \cos{x}\,dx \implies v = \sin{x} \end{align*}, then

\displaystyle \displaystyle \begin{align*} \int{\cos^2{x}\,dx} &= \sin{x}\cos{x} - \int{-\sin^2{x}\,dx} \\ \int{\cos^2{x}\,dx} &= \sin{x}\cos{x} + \int{\sin^2{x}\,dx} \\ \int{\cos^2{x}\,dx} &= \sin{x}\cos{x} + \int{1 - \cos^2{x}\,dx} \\ \int{\cos^2{x}\,dx} &= \sin{x}\cos{x} + \int{1\,dx} - \int{\cos^2{x}\,dx} \\ 2\int{\cos^2{x}\,dx} &= \sin{x}\cos{x} + x \\ \int{\cos^2{x}\,dx} &= \frac{1}{2}\sin{x}\cos{x} + \frac{1}{2}x + C \end{align*}

5. ## Re: Integrate cos^2 t using integration by parts

Hi (slow response on my part, sorry). Thanks for all your replies.

Yes, we had to use integration by parts. Integral of cos^2 was actually part of a larger trig substitution problem (which I'm not going to type up right now) where the teacher wanted them to "unsubstitute" using a triangle, and we couldn't figure out how to do that using the identity of pickslides response.

In other words, the whole problem was a trig substitution where we had to substitute (something like) x=3tan(t), and use use an identity to simplify sqrt(9-x^2) in the denominator and something else, blah blah. We resolved all the integrals, but were left with integral cos^2(t). At the end, the teacher wants them to unsubstitute by using tan(t)=x/3 and labeling the triangle's sides and figuring out how to "unsubstitute" (return to an equation in variable x). But the first trig identity leaves us with a cos(2t), and we do not see a way "unsubstitute" a trig function(2t) when all we have is tan(t)=x/3. I insisted on integration by parts because I wanted to keep the trig functions in t, rather than 2t.

6. ## Re: Integrate cos^2 t using integration by parts

Hi (slow response on my part, sorry). Thanks for all your replies.

Yes, we had to use integration by parts. Integral of cos^2 was actually part of a larger trig substitution problem (which I'm not going to type up right now) where the teacher wanted them to "unsubstitute" using a triangle, and we couldn't figure out how to do that using the identity of pickslides response.

In other words, the whole problem was a trig substitution where we had to substitute (something like) x=3tan(t), and use use an identity to simplify sqrt(9-x^2) in the denominator and something else, blah blah. We resolved all the integrals, but were left with integral cos^2(t). At the end, the teacher wants them to unsubstitute by using tan(t)=x/3 and labeling the triangle's sides and figuring out how to "unsubstitute" (return to an equation in variable x). But the first trig identity leaves us with a cos(2t), and we do not see a way "unsubstitute" a trig function(2t) when all we have is tan(t)=x/3. I insisted on integration by parts because I wanted to keep the trig functions in t, rather than 2t.
\displaystyle \displaystyle \begin{align*} \cos{2t} &\equiv 2\cos^2{t} - 1 \\ &\equiv \frac{2}{\sec^2{t}} - 1 \\ &\equiv \frac{2 - \sec^2{t}}{\sec^2{t}} \\ &\equiv \frac{2 - \left(1 + \tan^2{t}\right)}{1 + \tan^2{t}} \\ &\equiv \frac{1 - \tan^2{t}}{1 + \tan^2{t}} \end{align*}

ALL trigonometric functions can be written in terms of every other trigonometric function.

7. ## Re: Integrate cos^2 t using integration by parts

Originally Posted by Prove It
\displaystyle \displaystyle \begin{align*} \cos{2t} &\equiv 2\cos^2{t} - 1\end{align}

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Ah ha! I'll bet this was the way his teacher wanted him to answer. Thx

8. ## Re: Integrate cos^2 t using integration by parts

Ah ha! I'll bet this was the way his teacher wanted him to answer. Thx
What? It shows up perfectly fine for me...

9. ## Re: Integrate cos^2 t using integration by parts

Originally Posted by Prove It
What? It shows up perfectly fine for me...
We can see it perfectly in your post, it's just quoting that seems to destroy your beautiful masterpiece.

10. ## Re: Integrate cos^2 t using integration by parts

Originally Posted by Prove It
What? It shows up perfectly fine for me...
Fixed. I only wanted to quote part of your post and I screwed up the editing.

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