Q.Find $\int \dfrac{(3\sin\phi - 2)\cos\phi}{5 - \cos^2\phi - 4\sin\phi}\, d\phi$

Concept: U Substitution — the denominator simplifies to a quadratic in $\sin\phi$, so substitute $u = \sin\phi$.

Step 1: Let $u = \sin\phi$, so $du = \cos\phi\, d\phi$. The numerator becomes $(3u - 2)\cos\phi\, d\phi = (3u - 2)\, du$.

Step 2: Rewrite the denominator using $\cos^2\phi = 1 - \sin^2\phi = 1 - u^2$:

Step 3: The integral becomes

Perform polynomial division: $3u - 2 = 3(u - 2) + 4$, so

$\int \left( \frac{3}{u - 2} + \frac{4}{(u - 2)^2} \right) du = 3\log|u - 2| - \frac{4}{u - 2} + C.$ …

Substitute $t=\sin\phi,\ dt=\cos\phi\,d\phi$. Using $\cos^2\phi=1-t^2$:

So

Partial fractions. Write $\dfrac{3t-2}{(t-2)^2}=\dfrac{A}{t-2}+\dfrac{B}{(t-2)^2}$, so $3t-2=A(t-2)+B$, giving $A=3$ and $B=3(2)-2=4$:

$\int\left(\frac{3}{t-2}+\frac{4}{(t-2)^2}\right)dt=3\log|t-2|-\frac{4}{t-2}+C.$ …

Method: Trig-to-Algebra Substitution, Then Partial Fractions

Use this for a trigonometric integral whose numerator contains the derivative of a single trig function present throughout: substitute that function to obtain a rational integral.

Steps

Step 1: Substitute the recurring trig function.

Spot that $\cos\phi\,d\phi$ multiplies everything and the rest depends on $\sin\phi$. Set $t=\sin\phi$, $dt=\cos\phi\,d\phi$, and convert $\cos^2\phi=1-t^2$.

Step 2: Simplify the denominator. …

Common Mistakes

Mistake 1: Not replacing $\cos^2\phi$ with $1-\sin^2\phi$.

Why it's wrong: leaving $\cos^2\phi$ blocks the reduction to a rational function of $t$. Correct approach: use $\cos^2\phi=1-t^2$ so the denominator becomes $(t-2)^2$.

Mistake 2: Integrating $\dfrac{4}{(t-2)^2}$ as a logarithm. …