Find \int \frac{\sin t}{(4+\cos^2 t) (2-\sin^2 t)} dt.

Question:

Find {eq}\int \frac{\sin t}{(4+\cos^2 t) (2-\sin^2 t)} dt {/eq}.

Integration by Substitution and by Partial Fraction:

Integration by Substitution: The integration which uses the u-substitution or v-substitution or w-substitution is called the integral by substitution method.

Integration by Partial Fraction: The partial factor which is in the form of {eq}\displaystyle \frac{P}{pu+q} {/eq} is called a linear factor of a partial fraction.

The general integral rules:

1. Eject the constant out: {eq}\displaystyle \int a\cdot f\left(u\right)du=a\cdot \int f\left(u\right)du. {/eq}

2. The sum rule: {eq}\displaystyle \int f\left(u\right)\pm g\left(u\right)du=\int f\left(u\right)du\pm \int g\left(u\right)du. {/eq}

3. Common integration: {eq}\displaystyle \int \frac{1}{u^2+1}du=\arctan \left(u\right). {/eq}

4. Common derivative: {eq}\displaystyle \frac{d}{dx}\left(\cos \left(x\right)\right)=-\sin \left(x\right). {/eq}

Answer and Explanation:

We have to evaluate the integration of $$\displaystyle I = \int \frac{\sin t}{(4+\cos^2 t) (2-\sin^2 t)} dt $$

Apply the substitution for {eq}u=\cos \left(t\right) \Rightarrow du = -\sin \left(t\right)dt. {/eq}

$$\displaystyle = \int -\frac{1}{\left(4+u^2\right)\left(u^2+1\right)}du $$

Eject the constant out.

$$\displaystyle = -\int \frac{1}{\left(4+u^2\right)\left(u^2+1\right)}du $$

Now, we are going to find its partial fractions.

{eq}\displaystyle \frac{1}{\left(u^2+4\right)\left(u^2+1\right)}=\frac{a_1u+a_0}{u^2+4}+\frac{a_3u+a_2}{u^2+1} {/eq}

Simplify:

{eq}\displaystyle 1=\left(a_1u+a_0\right)\left(u^2+1\right)+\left(a_3u+a_2\right)\left(u^2+4\right) {/eq}

Expand:

{eq}\displaystyle 1=a_1u^3+a_1u+a_0u^2+a_0+a_3u^3+4a_3u+a_2u^2+4a_2 {/eq}

Group the elements accourding to powers of x.

{eq}\displaystyle 1=u^3\left(a_1+a_3\right)+u^2\left(a_0+a_2\right)+u\left(a_1+4a_3\right)+\left(a_0+4a_2\right) {/eq}

Equate the coefficients of similar terms on both sides to create a list of equations:

{eq}\displaystyle a_0+4a_2=1\\ \displaystyle a_1+4a_3=0\\ \displaystyle a_0+a_2=0\\ \displaystyle a_1+a_3=0. {/eq}

Now, solving above equations we get the value of {eq}a_3=0, a_1=0, a_2=\frac{1}{3} {/eq} and {eq}a_0=-\frac{1}{3}. {/eq}

So, partial fractions are given by

$$\displaystyle = -\int \frac{1}{3\left(u^2+1\right)}-\frac{1}{3\left(u^2+4\right)}du $$

Apply the sum rule.

$$\displaystyle = -\left(\int \frac{1}{3\left(u^2+1\right)}du-\int \frac{1}{3\left(u^2+4\right)}du\right) $$

Eject the constant out.

$$\displaystyle = -\left(\frac{1}{3}\cdot \int \frac{1}{u^2+1}du-\frac{1}{3}\cdot \int \frac{1}{u^2+4}du\right) $$

Apply the substitution for {eq}u=2v \Rightarrow du = 2dv. {/eq}

$$\displaystyle = -\left(\frac{1}{3}\cdot \int \frac{1}{u^2+1}du-\frac{1}{3}\cdot \int \frac{1}{2\left(v^2+1\right)}dv\right) $$

Eject the constant out.

$$\displaystyle = -\left(\frac{1}{3}\cdot \int \frac{1}{u^2+1}du-\frac{1}{3}\cdot \frac{1}{2}\cdot \int \frac{1}{v^2+1}dv\right) $$

Use the common integration.

$$\displaystyle = -\left(\frac{1}{3}\arctan \left(u\right)-\frac{1}{3}\cdot \frac{1}{2}\arctan \left(v\right)\right)+C $$

Substitute back {eq}v=\frac{u}{2}. {/eq}

$$\displaystyle = -\left(\frac{1}{3}\arctan \left(u\right)-\frac{1}{3}\cdot \frac{1}{2}\arctan \left(\frac{u}{2}\right)\right)+C $$

Simplify:

$$\displaystyle = -\left(\frac{1}{3}\arctan \left(u\right)-\frac{1}{6}\arctan \left(\frac{u}{2}\right)\right)+C $$

Substitute back {eq}u=\cos \left(t\right). {/eq}

$$\displaystyle = -\left(\frac{1}{3}\arctan \left(\cos \left(t\right)\right)-\frac{1}{6}\arctan \left(\frac{\cos \left(t\right)}{2}\right)\right)+C $$

Simplify:

$$\displaystyle = -\frac{1}{3}\arctan \left(\cos \left(t\right)\right)+\frac{1}{6}\arctan \left(\frac{1}{2}\cos \left(t\right)\right)+C $$

Where C is constant of the integration.


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