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1

### AIEEE 2006

$$\int\limits_0^\pi {xf\left( {\sin x} \right)dx}$$ is equal to
A
$$\pi \int\limits_0^\pi {f\left( {\cos x} \right)dx}$$
B
$$\,\pi \int\limits_0^\pi {f\left( {sinx} \right)dx}$$
C
$${\pi \over 2}\int\limits_0^{\pi /2} {f\left( {sinx} \right)dx}$$
D
$$\pi \int\limits_0^{\pi /2} {f\left( {\cos x} \right)dx}$$

## Explanation

$$I = \int\limits_0^\pi {xf\left( {\sin \,x} \right)dx}$$

$$= \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin x} \right)dx}$$

$$= \pi \int\limits_0^\pi {f\left( {\sin x} \right)dx - 1}$$

$$\Rightarrow 2I = \pi {\pi \over 0}f\left( {\sin x} \right)dx$$

$$I = {\pi \over 2}\int\limits_0^\pi {f\left( {\sin x} \right)dx}$$

$$= \pi \int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx}$$

$$= \pi \int\limits_0^{\pi /2} {f\left( {\cos x} \right)dx}$$
2

### AIEEE 2005

The value of integral, $$\int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx$$ is
A
$${1 \over 2}$$
B
$${3 \over 2}$$
C
$$2$$
D
$$1$$

## Explanation

$$I = \int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$I = \int\limits_3^6 {{{\sqrt {9 - x} } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

$$\left[ \, \right.$$ using $$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {f\left( {a + b - x} \right)dx} }$$ $$\left. \, \right]$$

Adding equation $$(1)$$ and $$(2)$$

$$2I = \int\limits_3^6 {dx} = 3 \Rightarrow I = {3 \over 2}$$
3

### AIEEE 2005

The value of $$\int\limits_{ - \pi }^\pi {{{{{\cos }^2}} \over {1 + {a^x}}}dx,\,\,a > 0,}$$ is
A
$$a\,\pi$$
B
$${\pi \over 2}$$
C
$${\pi \over a}$$
D
$${2\pi }$$

## Explanation

Let $$I = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {a^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$= \int\limits_{ - \pi }^\pi {{{{{\cos }^2}\left( { - x} \right)} \over {1 + {a^{ - x}}}}} dx$$

$$\left[ \, \right.$$ Using $$\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)dx}$$ $$\left. \, \right]$$

$$= \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {\alpha ^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

Adding equations $$(1)$$ and $$(2)$$ we get

$$2I = \int\limits_{ - \pi }^\pi {{{\cos }^2}} x\left( {{{1 + {a^x}} \over {1 + {a^x}}}} \right)dx$$

$$= \int\limits_{ - \pi }^\pi {{{\cos }^2}} x\,dx$$

$$= 2\int\limits_0^\pi {{{\cos }^2}} x\,dx$$

$$= 2 \times 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}} x\,dx$$

$$= 4\int\limits_0^{{\pi \over 2}} {{{\sin }^2}} x\,dx$$

$$\Rightarrow I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}} \,x\,dx$$

$$= 2\int\limits_0^{{\pi \over 2}} {\left( {1 - {{\cos }^2}x\,dx} \right)}$$

$$\Rightarrow I\,\, = 2\int\limits_0^{{\pi \over 2}} {dx - 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}} } \,x\,dx$$

$$\Rightarrow I + I = 2\left( {{\pi \over 2}} \right) = \pi$$

$$\Rightarrow I = {\pi \over 2}$$
4

### AIEEE 2005

Let $$f(x)$$ be a non - negative continuous function such that the area bounded by the curve $$y=f(x),$$ $$x$$-axis and the ordinates $$x = {\pi \over 4}$$ and $$x = \beta > {\pi \over 4}$$ is $$\left( {\beta \sin \beta + {\pi \over 4}\cos \beta + \sqrt 2 \beta } \right).$$ Then $$f\left( {{\pi \over 2}} \right)$$ is
A
$$\left( {{\pi \over 4} + \sqrt 2 - 1} \right)$$
B
$$\left( {{\pi \over 4} - \sqrt 2 + 1} \right)$$
C
$$\left( {1 - {\pi \over 4} - \sqrt 2 } \right)$$
D
$$\left( {1 - {\pi \over 4} + \sqrt 2 } \right)$$

## Explanation

Given that

$$\int\limits_{\pi /4}^\beta {f\left( x \right)} dx = \beta \sin \beta + {\pi \over 4}\cos \,\beta + \sqrt 2 \beta$$

Differentiating $$w.r.t$$ $$\beta$$

$$f\left( \beta \right) = \beta \cos \beta + \sin \beta - {\pi \over 4}\sin \beta + \sqrt 2$$

$$f\left( {{\pi \over 2}} \right) = \left( {1 - {\pi \over 4}} \right)\sin {\pi \over 2} + \sqrt 2$$

$$= 1 - {\pi \over 4} + \sqrt 2$$

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