Question

π/4∫ dx/2+3cos²x ,Evaluate it.0

Answer 1

Answer:

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Answer 2

Answer:

$\frac{1}{\sqrt{10}}$.tan¬1($\frac{\sqrt{2} }{\sqrt{5} }$)

Step-by-step explanation:

We have to evaluate, $\int\limits^\frac{\pi }{4} _0 {\frac{1}{2+3Cos^{2}x } } \, dx$.

Let us calculate the indefinite integral of the function first. Then we will put the limits to get the final answer.

∫$\frac{dx}{2+3Cos^{2}x }$

=∫$\frac{Sec^{2}x.dx }{2Sec^{2}x+3 }$

=∫$\frac{d(tanx)}{2(1+tan^{2}x)+3 }$

=∫$\frac{d(tanx)}{2tan^{2}x+5 }$

=$\frac{1}{2}$∫$\frac{d(tanx)}{tan^{2}x+\frac{5}{2}}$

=$\frac{1}{2}.\frac{\sqrt{2} }{\sqrt{5}}.$.tan¬1($\frac{\sqrt{2}tanx }{\sqrt{5} }$)+ c {Where c is an integration constant)

[Here, we have applied the formula of integration:

∫$\frac{dx}{x^{2}+a^{2}}$=$\frac{1}{a}$tan¬1($\frac{x}{a}$)]

Now, we will put the limits from x=0 to x=π/4 in this indefinite integral.

Hence,

$\int\limits^\frac{\pi }{4} _0 {\frac{1}{2+3Cos^{2}x } } \, dx$

=$\frac{1}{\sqrt{10}}$.tan¬1($\frac{\sqrt{2}.1 }{\sqrt{5} }$)-$\frac{1}{\sqrt{10}}$.tan¬1(0)

=$\frac{1}{\sqrt{10}}$.tan¬1($\frac{\sqrt{2} }{\sqrt{5} }$)

(Answer)