Integrate cos x/[(4+sin²x)(5−4cos²x)]

A standard Class 12 integration problem, not tied to a specific exam paper.

Problem

Evaluate:

$$\int \frac{\cos x}{(4+\sin^2 x)(5-4\cos^2 x)}\,dx$$

Key insight. The two factors in the denominator look unrelated at first — one in $\sin^2x$, one in $\cos^2x$ — but rewriting $\cos^2x$ as $1-\sin^2x$ turns both factors into expressions purely in $\sin x$. Combined with $dx$ pairing naturally with $\cos x\,dx$, that makes $t=\sin x$ the substitution the whole integral is built for.

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Approach

Rewrite $5-4\cos^2x$ purely in terms of $\sin x$, then substitute $t=\sin x$ so that $\cos x\,dx$ becomes $dt$. The integral reduces to a rational function of $t$ with two quadratic factors in the denominator, which splits by partial fractions into two standard arctangent integrals.

Solution

Step 1 — Rewrite the second factor in terms of $\sin x$

$$5 – 4\cos^2x = 5 – 4(1-\sin^2x) = 1+4\sin^2x$$

So the integral becomes:

$$\int \frac{\cos x}{(4+\sin^2x)(1+4\sin^2x)}\,dx$$

Step 2 — Substitute $t = \sin x$

With $dt = \cos x\,dx$:

$$\int \frac{dt}{(4+t^2)(1+4t^2)}$$

Step 3 — Split by partial fractions

Treating $t^2$ as a single variable $u$: write $\dfrac{1}{(4+u)(1+4u)} = \dfrac{A}{4+u}+\dfrac{B}{1+4u}$.

Clearing denominators: $1 = A(1+4u) + B(4+u)$.

Setting $u=-4$: $1 = A(1-16) = -15A \implies A = -\dfrac{1}{15}$

Setting $u=-\dfrac{1}{4}$: $1 = B\left(4-\dfrac{1}{4}\right) = \dfrac{15}{4}B \implies B = \dfrac{4}{15}$

So:

$$\frac{1}{(4+t^2)(1+4t^2)} = -\frac{1}{15}\cdot\frac{1}{4+t^2} + \frac{4}{15}\cdot\frac{1}{1+4t^2}$$

Step 4 — Integrate each term

Using $\displaystyle\int \frac{dt}{4+t^2} = \frac{1}{2}\tan^{-1}\left(\frac{t}{2}\right)$ and $\displaystyle\int \frac{dt}{1+4t^2} = \frac{1}{2}\tan^{-1}(2t)$:

$$\int \frac{dt}{(4+t^2)(1+4t^2)} = -\frac{1}{15}\cdot\frac{1}{2}\tan^{-1}\left(\frac{t}{2}\right) + \frac{4}{15}\cdot\frac{1}{2}\tan^{-1}(2t) + C$$

$$= -\frac{1}{30}\tan^{-1}\left(\frac{t}{2}\right) + \frac{2}{15}\tan^{-1}(2t) + C$$

Step 5 — Substitute back $t=\sin x$

Replacing $t$ with $\sin x$ throughout gives the final result in terms of the original variable.

Answer

Common mistakes

• Forgetting to convert $\cos^2x$ into $\sin^2x$ before substituting. Without that step, the two denominator factors mix $\sin x$ and $\cos x$, and $t=\sin x$ no longer cleanly absorbs both.
• Mismatching the standard arctangent form. $\displaystyle\int\frac{dt}{1+4t^2}$ is not $\tan^{-1}(2t)$ without the leading $\frac{1}{2}$ — treating $1+4t^2$ as $1+(2t)^2$ and remembering the $\frac{1}{a}\tan^{-1}(t/a)$ pattern (here with $a$ effectively absorbed into the substitution) avoids the slip.

Practise next

• Evaluate $\displaystyle\int \frac{\sin x}{(9+\cos^2x)(2-\cos^2x)}\,dx$ using the same $t=\cos x$ substitution and partial-fractions approach.