Q.Evaluate the definite integral $\int_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x}\,dx$

The numerator $\sin x+\cos x$ is the derivative of $\sin x-\cos x$, and $(\sin x-\cos x)^2=1-\sin 2x$. So substitute $t=\sin x-\cos x$.

Then $dt=(\cos x+\sin x)\,dx$ and $\sin 2x=1-t^2$. Limits: $x=0\Rightarrow t=-1$, $x=\tfrac{\pi}{4}\Rightarrow t=0$. The denominator becomes $9+16(1-t^2)=25-16t^2$:

With $25-16t^2=16\big((\tfrac54)^2-t^2\big)$ and $\int\frac{dt}{a^2-t^2}=\frac{1}{2a}\log\left|\frac{a+t}{a-t}\right|$: …

Spotting the substitution

Ask whose derivative is $\sin x+\cos x$. Since $\dfrac{d}{dx}(\sin x-\cos x)=\cos x+\sin x$, the quantity $t=\sin x-\cos x$ has precisely this numerator as its differential. Squaring it links it to the denominator:

This is the standard move when the numerator is $\sin x\pm\cos x$ and the denominator involves $\sin 2x$.

Change everything to $t$

$dt=(\sin x+\cos x)\,dx$ replaces the numerator times $dx$. Limits: at $x=0$, $t=0-1=-1$; at $x=\tfrac{\pi}{4}$, $t=\tfrac{\sqrt2}{2}-\tfrac{\sqrt2}{2}=0$. The denominator:

Hence

Evaluate the standard integral

Write $25-16t^2=16\left(\left(\tfrac54\right)^2-t^2\right)$, so with $a=\tfrac54$, …

Method: $t=\sin x-\cos x$ reducing a $\sin 2x$ denominator to $\int\frac{dt}{a^2-t^2}$

Use this when the numerator is $\sin x\pm\cos x$ and the denominator is a constant plus a multiple of $\sin 2x$: the substitution converts it into a standard $a^2-t^2$ form.

Steps

Step 1: Substitute $t=\sin x-\cos x$.

Then $dt=(\sin x+\cos x)\,dx$ (the numerator) and $\sin 2x=1-t^2$.

Step 2: Rewrite the denominator. …

Common Mistakes

Mistake 1: Not linking the numerator to $d(\sin x-\cos x)$.

Why it's wrong: $\sin x+\cos x$ is exactly the derivative of $\sin x-\cos x$; missing this hides the substitution $t=\sin x-\cos x$. Correct approach: set $t=\sin x-\cos x$.

Mistake 2: Wrong standard formula for $\int\dfrac{dt}{a^2-t^2}$. …