Question

Evaluate the indefinite integral. (use c for the constant of integration.) sin(2x) 24 + cos2(x) dx

Answer 1

Possibly you refer to the integral,

$\displaystyle\int\frac{\sin2x}{24+\cos^2x}\,\mathrm dx$

Recall the double angle identity,

$\cos^2x=\dfrac{1+\cos2x}2$

Then the integral is equivalent to

$\displaystyle\int\frac{\sin2x}{24+\frac{1+\cos2x}2}\,\mathrm dx=2\int\frac{\sin2x}{49+\cos2x}\,\mathrm dx$

Let $y=49+\cos2x$ so that $\mathrm dy=-2\sin2x\,\mathrm dx$. Then the integral becomes

$\displaystyle-\int\frac{\mathrm dy}y=-\ln|y|+C$

$\implies\displaystyle\int\frac{\sin2x}{24+\cos^2x}\,\mathrm dx=-\ln|49+\cos2x|+C$

We can rewrite this to try to get it in a form that more resembles the original integrand.

$-\ln|49+\cos2x|+C=-\ln\left|2\left(24+\dfrac{1+\cos2x}2\right)\right|+C$

$=-\ln2-\ln\left|24+\cos^2x\right|+C$

$=-\ln\left|24+\cos^2x\right|+C$

where the constant term $-\ln2$ got absorbed into the general constant $C$.

Answer 2

Using substitution, the result of the integral is given by:

$-\ln{(49 + \cos{2x})} + C$

We want to calculate the following integral:

$\int \frac{\sin{2x}}{24 + \cos^{2}x} dx$

The following identity is used to simplify the cosine squared:

$\cos^2{x} = \frac{1 + \cos{2x}}{2}$

Replacing into the integral:

$\int \frac{\sin{2x}}{24 + \frac{1 + \cos{2x}}{2}} dx$

$\int \frac{2\sin{2x}}{48 + 1 + \cos{2x}} dx$

$\int \frac{2\sin{2x}}{49 + \cos{2x}} dx$

The substitution is:

$u = 49 + \cos{2x}$

$du = -2\sin{2x}dx$

$dx = -\frac{1}{2\sin{2x}} du$

Hence:

$\int \frac{2\sin{2x}}{49 + \cos{2x}} dx$

$\int \frac{2\sin{2x}}{u} -\frac{1}{2\sin{2x}} du$

$\int -\frac{1}{u} du$

$-\ln{u} + C$

Since $u = 49 + \cos{2x}$, the result as a function of x is:

$-\ln{(49 + \cos{2x})} + C$

A similar problem is given at https://brainly.com/question/22077165