Question

${\int }_0^{\frac{\pi }{2}}\frac{dx}{a^{2}co{s}^{2}x+b^{2}si{n}^{2}x}$ where (a, b >0) is equal to-

• A. $\frac{\pi }{2ab}$
• B. $\frac{2}{\pi ab}$
• C. $\frac{\pi }{2a}$
• D. $\frac{\pi }{2b}$

Answer 1

The correct option is A $\frac{\pi }{2ab}$

${\int }_0^{\frac{\pi }{2}}\frac{dx}{a^{2}co{s}^{2}x+b^{2}si{n}^{2}x}$
$={\int }_0^{\frac{\pi }{2}}\frac{se{c}^{2}xdx}{a^2+b^{2}ta{n}^{2}x}$(on dividing numerator and denominator by $co{s}^{2}x)$
Put $tanx=t\Rightarrow se{c}^{2}xdx=dt$
On substituting tanx = t we’ll have our integrand as $\frac{dt}{a^2+b^2{t}^2}$
But what’ll be the limits ? Will they be the same?
No, they’ll change as we substitute tanx = t. Limits of definite integral is the limit of the variable of which infinitesimally small part is present in the integrand (Like it’ll be of “x” if dx is there, and of “t” if dt is there etc.)
The given limits in the question from $(0,\frac{\pi }{2})$ are the limits of “x”. On substituting tanx = t we’ll have dt in the integrand and not dx. So limits will change according to “t”.
$tanx=t\Rightarrow t=0$ when x = 0
And $t=\infty$ when x = $\frac{\pi }{2}$
So, the limits will be $(0,\infty )$
Therefore,
${\int }_0^{\frac{\pi }{2}}\frac{dx}{a^{2}co{s}^{2}x+b^{2}si{n}^{2}x}={\int }_0^{\infty }\frac{dt}{a^2+b^2{t}^2}$
$=\frac{1}{b^2}{\int }_0^{\infty }\frac{dt}{(\frac{a}{b}{)}^2+t^2}$
$=(\frac{1}{b^2}.\frac{1}{a/b}[ta{n}^{-1}\left(\frac{bt}{a}\right)\left]\right){|}_0^{\infty }$
$=\frac{1}{a.b}[\frac{\pi }{2}-0]$
$=\frac{\pi }{2ab}$