Question

$\int \frac{1}{2+3sinx}dx$

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

The above problem is of the form $\int \frac{1}{a+bsinx}dx$ and we saw that, to integrate these kind of forms we’ll use the half angle formula of sinx in terms of $tan\left(\frac{x}{2}\right)$

Which is $sinx=2\frac{tan\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}$

We’ll substitute $tan\frac{x}{2}=t$

Let’s use the same method here

So, the given integral would be = $\int \frac{1}{2+3.2\frac{tan\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}}dx$

$=\int \frac{1+ta{n}^2\left(\frac{x}{2}\right)}{2(1+ta{n}^2\frac{x}{2})+6tan\frac{x}{2}}dx=\int \frac{se{c}^2\left(\frac{x}{2}\right)}{2(1+ta{n}^2\frac{x}{2})+6tan\frac{x}{2}}dx$

Let’s substitute $tan\frac{x}{2}$ = t

We get $\frac{1}{2}se{c}^2\left(\frac{x}{2}\right).dx=dt$

So, we’ll have the integrals in terms of t -

$\int \frac{2}{2+2t^2+6t}dtOr\int \frac{1}{1+t^2+3t}dt$

We can see that the integral we have now has a quadratic expression in the denominator. We have solved these kind of integrals earlier also by making the quadratic expression a perfect square. We’ll do the same here -

$\int \frac{1}{t^2+3t+\frac{9}{4}-\frac{5}{4}}dtOr\int \frac{1}{{(t+\frac{3}{2})}^2-\frac{5}{4}}dt$

This is in the form of $\int \frac{1}{x^2-a^2}dx$

And we can directly use the corresponding formula. Which is -

$\int \frac{1}{x^2-a^2}dx=\frac{1}{2a}log\left|\frac{x-a}{x+a}\right|+C$

On using this formula we’ll get

$\frac{\sqrt{4}}{2\sqrt{5}.}log\left|\frac{t+\frac{3}{2}-\sqrt{\frac{5}{4}}}{t+\frac{3}{2}+\sqrt{\frac{5}{4}}}\right|+C$

Let’s substitute the value of t in the above expression.

$\frac{1}{\sqrt{5}}.log\left|\frac{tan\frac{x}{2}+\frac{3}{2}-\sqrt{\frac{5}{4}}}{tan\frac{x}{2}+\frac{3}{2}+\sqrt{\frac{5}{4}}}\right|+COr\frac{1}{\sqrt{5}}.log\left|\frac{tan\frac{x}{2}+\frac{3}{2}-\sqrt{\frac{5}{4}}}{tan\frac{x}{2}+\frac{3}{2}+\sqrt{\frac{5}{4}}}\right|+C$