Question

π/4
∫ dx/a² cos²x-b² sin²x (a > b > 0) ,Evaluate it.
0

Answer 1

Answer:

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Answer 2

Answer:

$\mathbf{\frac{1}{2ab}log\frac{a-b}{a+b}}$

Step-by-step explanation:

In this question,

We have to evaluate

$\int\limits^\frac{\pi}{4}_0 {\frac{1}{a^{2}cos^{2}x\ -\ b^{2}sin^{2}x} \, dx$

Multiplying by sec²x in numerator and denominator we get,

$\int\limits^\frac{\pi}{4}_0 {\frac{sec^{2}x}{a^{2}sec^{2}xcos^{2}x\ -\ b^{2}sec^{2}xsin^{2}x} \, dx$

$\int\limits^\frac{\pi}{4}_0 {\frac{sec^{2}x}{a^{2}\ -\ b^{2}tan^{2}x} \, dx$

Since $sec^{2}x\ =\ \frac{1}{cos^{2}x}$

Now, on substituting tanx = t we get

sec²xdx = dt

Putting the values and taking $\frac{1}{b^{2}}$ we get

$\frac{1}{b^{2}}\int\limits^1_0{\frac{1}{\frac{a^{2}}{b^{2}}\ -\ t^{2}}} \, dt$

$\frac{1}{2ab}^{1}_{0}[{log\frac{\frac{a}{b}-t}{\frac{a}{b}+t}]$

Using upper and lower limits we get,

$\mathbf{\frac{1}{2ab}log\frac{a-b}{a+b}}$