How do I solve the following integral:

$\int (sin\theta + 3)(cos^2\theta) $

The next line in the solution reads that this is equal to

$ -1/3cos^3\theta + 3/2 \int cos2\theta + 1 $

However, I am not sure how they got there.

  • $\begingroup$ @MathematicsStudent1122 Why not make that an answer? $\endgroup$ – Cameron Williams Jul 10 '16 at 0:25
  • $\begingroup$ You forgot $d\theta$. $\endgroup$ – user236182 Jul 10 '16 at 0:31

Hint: After expanding, split the integral. For one of them, you have an obvious substitution; for the other, write $\cos^2 \theta = \frac 12 \cos 2\theta + \frac 12$; this is a well known trigonometric identity.


$=\int (\sin{\theta}+3)(1-\sin^2{\theta})$ $d\theta$

Expand. Use Identity

$sin^3{\theta}=\frac{1}{4}(3 \sin {\theta}-\sin{3\theta})$

Integral shall become

$=\int \sin{\theta}+3-\frac{1}{4}(3\sin{\theta}-\sin{3\theta})-\frac{3}{2}(1-\cos{2\theta})$ $d\theta$

$=-\cos{\theta}+3\theta-\frac{\cos{3\theta}}{12}+\frac{3}{4}\cos{\theta}+\frac{3}{2}\int(\cos{2\theta}-1)$ $d\theta$

Rewritng to get your above answer, i.e sending $3\theta$ in the integral.

$=-\frac{3\cos{\theta}}{12}-\frac{\cos{3\theta}}{12}+\frac{3}{2}\int(\cos{2\theta}-1+2)$ $d\theta$

$=-\frac{1}{3}[\frac{1}{4}(\cos{3\theta}+3\cos\theta)]+\frac{3}{2}\int(\cos{2\theta}+1)$ $d\theta$

Use Identity for $\cos^3{\theta}$

$=-\frac{1}{3}\cos^3{\theta}+\frac{3}{2}\int(\cos{2\theta}+1)$ $d\theta$

Hence the above.


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