Question

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Evaluate the following ! :)

$\Large {\displaystyle \sf \int \limits^{\pi}_ {0} \footnotesize \dfrac{x\ dx}{a^2\ cos^2\ x\ +\ b^2\ sin^2\ x}}$

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Answer 1

Step-by-step explanation:

We have,

$I= \int \limits^{\pi} _{0} \frac{xdx}{ {a}^{2} \cos ^{2} (x) + {b}^{2} \sin^{2} (x) } ...(i)$

$I= \int \limits^{\pi} _{0} \frac{(\pi - x)dx}{ {a}^{2} \cos ^{2} (\pi - x) + {b}^{2} \sin^{2} (\pi - x) }$

$I= \int \limits^{\pi} _{0} \frac{(\pi - x)dx}{ {a}^{2} \cos ^{2} (x) + {b}^{2} \sin^{2} (x) } ....(ii)$

Adding i and ii,

$2I= 2\pi\int \limits^{ \frac{\pi}{2}} _{0} \frac{dx}{ {a}^{2} \cos ^{2} (x) + {b}^{2} \sin^{2} (x) }$

Dividing num. and deno. by cos²(x),

$I= \pi\int \limits^{ \frac{\pi}{2}} _{0} \frac{ \sec^{2} (x) dx}{ {a}^{2} + {b}^{2} \tan^{2} (x) }$

$Let \tan(x) = t \implies \sec^{2}(x) dx= dt$

$I= \pi\int \limits^{ \infty } _{0} \frac{ dt}{ {a}^{2} +( b t)^{2} }$

$I= \pi .\frac{1}{ab} [ \tan^{ - 1} ( \frac{bt}{a} ) ]^{ \infty } _{0}$

$I= \frac{\pi}{ab} [ \tan^{ - 1} ( \infty ) - \tan ^{ - 1} (0) ]$

$I= \frac{\pi}{ab} . \frac{\pi}{2}$

$I= \frac{\pi^{2} }{2ab}$

Answer 2

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the value of the given integral is zero.

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To evaluate the given integral, let's go through the steps:

Step 1: Simplify the integrand.

The integrand can be simplified using the trigonometric identity:

$\cos^2(x) + \sin^2(x) = 1$

Using this identity, we can rewrite the integrand as:

$\frac{x}{a^2\cos^2(x) + b^2\sin^2(x)} = \frac{x}{a^2(1 - \sin^2(x)) + b^2\sin^2(x)}$

Simplifying further:

$\frac{x}{a^2 - a^2\sin^2(x) + b^2\sin^2(x)} = \frac{x}{a^2 + (b^2 - a^2)\sin^2(x)}$

Step 2: Use a trigonometric substitution.

Let's make a substitution by setting:

$u = \sin(x)$

Then, $du = \cos(x)dx$

We can rewrite the integral in terms of $u$:

$\int \frac{x}{a^2 + (b^2 - a^2)\sin^2(x)} dx = \int \frac{x}{a^2 + (b^2 - a^2)u^2} \frac{du}{\cos(x)}$

Step 3: Apply the limits of integration.

As the original limits of integration are $0$ and $\pi$, we need to express the new limits of integration in terms of $u$.

When $x = 0$, $u = \sin(0) = 0$

When $x = \pi$, $u = \sin(\pi) = 0$

Thus, the new limits of integration are from $0$ to $0$.

Step 4: Evaluate the integral.

Since the new limits of integration are the same, the integral evaluates to zero:

$\int_{0}^{0} \frac{x}{a^2 + (b^2 - a^2)u^2} \frac{du}{\cos(x)} = 0$

Therefore, the value of the given integral is zero

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