Q.Evaluate : $\int_{0}^{\pi/4} \dfrac{\sin x + \cos x}{16 + 9\sin 2x}\, dx$ OR Evaluate : $\int_{1}^{3} (x^2 + 3x + e^x)\, dx$ as the limit of the sum.

Step 1 — Substitute

Let $u = x+6$, so $du = dx$.

When $x = a-6$, $u=a$; when $x=b-6$, $u=b$.

Step 2 — Rewrite the integral …

The key insight is that the integrand $\frac{(x - x^3)^{1/3}}{x^4}$ looks like it might be related to trigonometric identities. The expression $x - x^3 = x(1 - x^2)$ suggests the substitution $x = \sin\theta$, because then $1 - x^2 = \cos^2\theta$ and the cube root becomes $(\sin\theta \cos^2\theta)^{1/3}$. The denominator $x^4 = \sin^4\theta$ then combines nicely, and the limits transform from $x = 1/3$ to $x = 1$ into $\theta = \arcsin(1/3)$ to $\theta = \pi/2$. After simplification, a further substitution $t = \cos\theta$ yields a straightforward integral.

1. Substitute $x = \sin\theta$.

   Then $dx = \cos\theta \, d\theta$.

   The limits: when $x = 1/3$, $\theta = \arcsin(1/3)$; when $x = 1$, $\theta = \pi/2$.

2. Rewrite the integrand.

So $(x - x^3)^{1/3} = (\sin\theta \cos^2\theta)^{1/3} = \sin^{1/3}\theta \cdot \cos^{2/3}\theta$.

The denominator: $x^4 = \sin^4\theta$.

Also $dx = \cos\theta \, d\theta$.

The integral becomes:

3. Simplify the exponent. Notice $\frac{\cos^{5/3}\theta}{\sin^{11/3}\theta} = \cot^{5/3}\theta \cdot \csc^2\theta$? Actually better: write as

But $\csc^2\theta \, d\theta = -\frac{d(\cot\theta)}{?}$ Wait: $d(\cot\theta) = -\csc^2\theta \, d\theta$. So $\csc^2\theta \, d\theta = -d(\cot\theta)$.

4. Substitute $t = \cot\theta$.

   Then $dt = -\csc^2\theta \, d\theta$.

   Limits: when $\theta = \arcsin(1/3)$, we need $\cot\theta$.

   If $\sin\theta = 1/3$, then $\cos\theta = \sqrt{1 - 1/9} = \sqrt{8/9} = \frac{2\sqrt{2}}{3}$.

   So $\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{2\sqrt{2}/3}{1/3} = 2\sqrt{2}$.

   When $\theta = \pi/2$, $\cot(\pi/2) = 0$.

   The integral becomes:

   $I = \int_{t = 2\sqrt{2}}^{0} t^{5/3} \cdot (-dt) = \int_{0}^{2\sqrt{2}} t^{5/3} \, dt.$ …

Step 1 — Spot the right substitution

The power of $\cos\theta$ is odd. That is the signal to save one factor of $\cos\theta$ (it will become $ds$) and convert the remaining even power using the Pythagorean identity:

Step 2 — Substitute

Let

Change the limits (essential for a definite integral):

The integral becomes

(using $\sqrt{s}\cdot s^{2}=s^{1/2+2}=s^{5/2}$).

Step 3 — Integrate term by term

Using $\displaystyle\int s^{n}\,ds=\frac{s^{n+1}}{n+1}$: …

The key idea here is that when you have a rational function with distinct linear factors in the denominator, the method of partial fractions lets you break it into simpler pieces — each of which integrates directly to a logarithm. The constants $A$, $B$, $C$ are exactly the numerators of those partial fractions.

Let’s work through it step by step.

1. Set up the partial fraction decomposition. Since the denominator $(x-1)(x-2)(x-3)$ has three distinct linear factors, we can write:

where $A$, $B$, $C$ are constants we need to find.

2. Clear the denominators. Multiply both sides by $(x-1)(x-2)(x-3)$:

3. Solve for the constants — the smart way.

   Instead of expanding everything and comparing coefficients (which works but is slower), we can substitute the roots of the denominator one by one. This is faster and less error-prone.

   • For $A$: Put $x=1$. Then the terms with $B$ and $C$ vanish because they contain $(x-1)$:

 So $A = 2$.

• For $B$: Put $x=2$:

 So $B = -7$.

• For $C$: Put $x=3$:

 So $C = 5$.

4. Verify quickly (optional but good practice). …