# How Do I Evaluate This Definite Integral?

$$\int_0^{\pi/2}\color{red}{3}\;\cos^2\theta\;d\theta$$

I tried doing it using $u$-substitution and then getting the anti-derivative. All in all, I keep getting $-1$, which is apparently wrong. I'd appreciate any help.

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That integral evaluates to a number. It is not a function, unless you are thinking of it as a constant function of some unnamed variable. So either there is nothing to differentiate, or you are differentiating a constant function, and in the latter case the answer is $0$. As for the integral itself, the value is $3\pi/4$. The identity $\cos(2\theta)=2\cos^2(\theta)-1$ is helpful for evaluating it. –  Jonas Meyer Feb 27 '12 at 4:44
Note that the value of the integral cannot possibly be $-1$: the function $f(\theta)=3\cos^2\theta$ is never negative, so the net area under the curve cannot be negative. –  Brian M. Scott Feb 27 '12 at 5:11

Hint: Use the double-angle formula $\cos 2\theta=2\cos^2\theta-1$. So $\cos^2\theta =\frac{\cos 2\theta +1}{2}$.

I believe you mean to ask how does one evaluate, or compute the definite integral.

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Thanks! That did the trick. –  user754950 Feb 27 '12 at 4:54

There is a cute trick for this:

$$I = 3\int_{0}^{\pi/2} \cos^2 x \text{ dx} = 3\int_{0}^{\pi/2} \sin^2 \text{ dx}$$

by using $\int_{a}^{b} f(x) \text{ dx} = \int_{a}^{b} f(a+b-x) \text{ dx}$

$$2I = 3 \int_{0}^{\pi/2} (\cos^2 x + \sin^2 x) \text{ dx} = 3 \int_{0}^{\pi/2} 1 \text{ dx} = \frac{3\pi}{2}$$

$$\frac{3\pi}{4}$$

But, there are more general methods which can be used. See the answers here: Evaluating $\int P(\sin x, \cos x) \text{d}x$

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$$\int_0^\frac{\pi}{2} 3 \cos^2\theta d\theta= \int_0^\frac{\pi}{2} \frac{3}{2}(1+\cos2\theta)d\theta$$ $$=\int_0^\frac{\pi}{2} \frac{3}{2}+\frac{3}{2}\cos 2\theta d\theta$$ $$=\frac{3}{2}(\frac{\pi}{2}-0)+\frac{3}{4}(\sin2\frac{\pi}{2}-\sin 0)$$ $$=\frac{3}{4}$$

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Just for fun, you can use integration by parts to find the antiderivative of $\cos^2 x$: \eqalign{ \color{maroon}{\int\cos^2 x\,dx} =\int\underbrace{\cos x}_u\,\underbrace{\cos x\,dx}_{dv}&= \underbrace{\cos x\vphantom{)}}_u\,\underbrace{( \sin x )}_{ v}- \int\underbrace{\sin x\vphantom{)}}_v\,\underbrace{(-\sin x\,dx)}_{du}\cr &=\sin x\cos x+\int\sin^2 x\,dx\cr &=\sin x\cos x+\int(1-\cos^2 x)\,dx\cr &=\sin x\cos x+ x-\color{maroon}{\int\cos^2 x \,dx};\cr } Whence $$\int\cos^2 x\,dx={\sin x\cos x+x\over 2}+C.$$

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