How do you use partial fractions to find the integral $int (sinx)/(cosx+cos^2x)dx$ ?

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Answer 1 · 2016-12-21T18:18:23

$ln|cosx + 1| - ln|cosx| + C$

We start with a substitution. Let $u = cosx$ . Then $du = -sinxdx -> dx = (du)/-sinx$ .

$=>intsinx/(u + u^2) xx (du)/-sinx$

$=> int -1/(u + u^2)du$

We now factor the denominator to $u(1 + u)$ . Let's now write $-1/(u + u^2)$ in partial fractions.

$A/u + B/(u + 1) = -1/(u(u +1))$

$A(u + 1) + B(u) = -1$

$Au + A + Bu = -1$

$(A + B)u + A = -1$

Now write a system of equations:

${(A + B = 0), (A = -1):}$

Solving, we get: $A = -1$ and $B = 1$ .

Thus, the partial fraction decomposition is $-1/u + 1/(u + 1)$ . The integral becomes $int(1/(u + 1) - 1/u)du$ .

We can now integrate using the rule $int(1/u)du = ln|u| + C$ .

$=> ln|u + 1| - ln|u| + C$

Finally, reinsert the value of $u$ to make the function defined by $x$ instead of $u$ :

$=>ln|cosx + 1| - ln|cosx| +C$

Hopefully this helps!

How do you use partial fractions to find the integral int (sinx)/(cosx+cos^2x)dxint (sinx)/(cosx+cos^2x)dx?