# Integral by use of substitution (int 3cos x dx)/(sqrt(1+3sin x))

Integral by use of substitution $\frac{3\mathrm{cos}xdx}{\sqrt{1+3\mathrm{sin}x}}$
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Eve Good
$\int \frac{3\mathrm{cos}\left(x\right)dx}{\sqrt{1+3\mathrm{sin}\left(x\right)}}$
Note that the derivative of $3\mathrm{sin}\left(x\right)+1$ is present $\left[3\mathrm{cos}\left(x\right)\right]$, so we can try to use u-substitution.
$u=1+3\mathrm{sin}\left(x\right)←$ what's inside the $\sqrt{}$
$du=3\mathrm{cos}\left(x\right)dx←$ precisely our numerator
After substitution, we have
$\int \frac{3\mathrm{cos}\left(x\right)dx}{\sqrt{1+3\mathrm{sin}\left(x\right)}}=\int \frac{1}{\sqrt{u}}du$
$\int \frac{3\mathrm{cos}\left(x\right)dx}{\sqrt{1+3\mathrm{sin}\left(x\right)}}=\int {u}^{-\frac{1}{2}}du$
Integrating
$\int \frac{3\mathrm{cos}\left(x\right)dx}{\sqrt{1+3\mathrm{sin}\left(x\right)}}=2{u}^{\frac{1}{2}}+C$
$\int \frac{3\mathrm{cos}\left(x\right)dx}{\sqrt{1+3\mathrm{sin}\left(x\right)}}=2\sqrt{u}+C$
Writing in terms of x
$\int \frac{3\mathrm{cos}\left(x\right)dx}{\sqrt{1+3\mathrm{sin}\left(x\right)}}=2\sqrt{1+3\mathrm{sin}\left(x\right)}+C$
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agantisbz
$\int \frac{3\mathrm{cos}x}{\sqrt{1+3\mathrm{sin}x}}dx=\int \frac{1}{\sqrt{u}}du$
take $u=1+3\mathrm{sin}x$
$du=3\mathrm{cos}xdx=\int {u}^{-1/2}du={u}^{1/2}+c=\sqrt{1+3\mathrm{sin}x}+c$