Evaluate the following:

(i) $\displaystyle\int_0^{\pi/2}\dfrac{dx}{1+5\cos^2x}$

(ii) $\displaystyle\int_0^{\pi/2}\dfrac{dx}{5+4\sin^2x}$

Question

Evaluate the following:

(i) $\displaystyle\int_0^{\pi/2}\dfrac{dx}{1+5\cos^2x}$

(ii) $\displaystyle\int_0^{\pi/2}\dfrac{dx}{5+4\sin^2x}$

Accepted Answer

> [!TLDR]
>
> Both integrals look proper (finite interval $[0,\pi/2]$) but $\sec x	o\infty$ at $x=\pi/2$; dividing by $\cos^2x$ and substituting $u=	an x$ reveals each as an improper integral in disguise, since $	an x	o\infty$ as $x	o(\pi/2)^-$.

**Step 1. (i) Divide the integrand by $\cos^2x$.** $\dfrac{1+5\cos^2x}{\cos^2x}=\sec^2x+5=(1+	an^2x)+5=	an^2x+6$, so
$$\dfrac{dx}{1+5\cos^2x}=\dfrac{\sec^2x\,dx}{	an^2x+6}.$$

**Step 2. (i) Substitute $u=	an x$, $du=\sec^2x\,dx$.** Limits: $x=0\Rightarrow u=0$; as $x	o(\pi/2)^-,\ u	o\infty$. So
$$\int_0^{\pi/2}\dfrac{dx}{1+5\cos^2x}=\int_0^\infty\dfrac{du}{u^2+6}.$$

**Step 3. (i) Evaluate the improper integral.** Using $\displaystyle\int\dfrac{du}{u^2+a^2}=\dfrac1a	an^{-1}\dfrac{u}{a}$ with $a=\sqrt6$, and $\lim_{t	o\infty}	an^{-1}t=\pi/2$:
$$\int_0^\infty\dfrac{du}{u^2+6}=\left[\dfrac1{\sqrt6}	an^{-1}\dfrac{u}{\sqrt6}ight]_0^\infty=\dfrac1{\sqrt6}\left(\dfrac{\pi}2-0ight)=\dfrac{\pi}{2\sqrt6}.$$

**Step 4. (ii) Divide the integrand by $\cos^2x$.** $\dfrac{5+4\sin^2x}{\cos^2x}=5\sec^2x+4	an^2x=5(1+	an^2x)+4	an^2x=5+9	an^2x$, so
$$\dfrac{dx}{5+4\sin^2x}=\dfrac{\sec^2x\,dx}{5+9	an^2x}.$$

**Step 5. (ii) Substitute $u=	an x$.** Same limits as before ($0	o\infty$):
$$\int_0^{\pi/2}\dfrac{dx}{5+4\sin^2x}=\int_0^\infty\dfrac{du}{9u^2+5}=\dfrac19\int_0^\infty\dfrac{du}{u^2+5/9}.$$

**Step 6. (ii) Evaluate with $a^2=5/9,\ a=\dfrac{\sqrt5}{3}$.** …

Q.Evaluate the following:

Concept understanding — Improper Integrals

Q.Evaluate the following:

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