Evaluate: $\displaystyle\int \frac{dx}{2\cos^2 x + 3\cos x}$.

Question

Accepted Answer

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> [!TLDR]
> Use the Weierstrass substitution $t=	an(x/2)$ to turn the trig integral into a rational-function integral, then apply partial fractions.
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> $\displaystyle\int\dfrac{dx}{2\cos^2x+3\cos x} = \int\dfrac{dx}{\cos x\,(2\cos x+3)}$
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> **Substitute $t=	an\dfrac{x}{2}$:** $\cos x = \dfrac{1-t^2}{1+t^2}$, $dx=\dfrac{2\,dt}{1+t^2}$.
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> Denominator: $\cos x(2\cos x+3) = \dfrac{1-t^2}{1+t^2}\cdot\dfrac{2(1-t^2)+3(1+t^2)}{1+t^2} = \dfrac{(1-t^2)(5+t^2)}{(1+t^2)^2}$
>
> So the integral becomes:
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> $\displaystyle\int \dfrac{2\,dt/(1+t^2)}{(1-t^2)(5+t^2)/(1+t^2)^2} = 2\int\dfrac{(1+t^2)\,dt}{(1-t^2)(5+t^2)}$
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> **Partial fractions** (using $1-t^2=(1-t)(1+t)$):
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> $\dfrac{1+t^2}{(1-t)(1+t)(5+t^2)} = \dfrac{1/6}{1-t}+\dfrac{1/6}{1+t}+\dfrac{-2/3}{5+t^2}$
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> (found by matching coefficients — the $t$-term coefficient vanishes automatically, and the constant/$t^2$ equations solve to $A=B=	frac16$, $D=-	frac23$.)
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> So the integral is:
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> $2\displaystyle\int\left[\dfrac{1/6}{1-t}+\dfrac{1/6}{1+t}-\dfrac{2/3}{5+t^2}ight]dt$
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> $= \dfrac13\Big(-\ln|1-t|\Big) + \dfrac13\ln|1+t| - \dfrac43\cdot\dfrac{1}{\sqrt5}	an^{-1}\!\left(\dfrac{t}{\sqrt5}ight) + C$
> …

Q.Evaluate: $\displaystyle\int \frac{dx}{2\cos^2 x + 3\cos x}$ .

Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

When it applies

The standard forms

Worked example

Q.Evaluate: ∫dx2cos⁡2x+3cos⁡x\displaystyle\int \frac{dx}{2\cos^2 x + 3\cos x}∫2cos2x+3cosxdx​.

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Q.Evaluate: $\displaystyle\int \frac{dx}{2\cos^2 x + 3\cos x}$ .

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