If -xcos- -1x2-2xcos-13-2 dx -fx-gx-1-C- then gx-2cos- fx is where C is constant of integration

Let I= ∫xcosα +1(x^2+2xcosα+1)^3/2 dx Now, ⇒ I =∫xcosα +1x^3(1+2xcosα+1x^2)^3/2 dx ⇒ I=∫cosαx^2+1x^3(1+2xcosα+1x^2)^3/2 dx Taking 1+2xcosα+1x^2=t^2 ⇒ 2cosαx ...

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Standard XIII

Mathematics

Integration by Substitution

If ∫xcosα +1x...

Question

If $\int \frac{x cos α + 1}{(x^{2} + 2 x cos α + 1)^{3 / 2}} d x = \frac{f (x)}{\sqrt{g (x) + 1}} + C$ , then $g (x) - 2 cos α f (x)$ is
(where $C$ is constant of integration)

x^{2}

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x

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2 x^{2}

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2 x

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Solution

The correct option is A $x^{2}$
Let $I = \int \frac{x cos α + 1}{(x^{2} + 2 x cos α + 1)^{3 / 2}} d x$
Now,
$\Rightarrow I = \int \frac{x cos α + 1}{x^{3} {(1 + \frac{2}{x} cos α + \frac{1}{x^{2}})}^{3 / 2}} d x \Rightarrow I = \int \frac{\frac{cos α}{x^{2}} + \frac{1}{x^{3}}}{{(1 + \frac{2}{x} cos α + \frac{1}{x^{2}})}^{3 / 2}} d x$

Taking $1 + \frac{2}{x} cos α + \frac{1}{x^{2}} = t^{2}$
$\Rightarrow - \frac{2 cos α}{x^{2}} - \frac{2}{x^{3}} = 2 t d t \Rightarrow \frac{cos α}{x^{2}} + \frac{1}{x^{3}} = - t d t$

Now,
$I = - \int \frac{t}{t^{3}} d t \Rightarrow I = - \int \frac{1}{t^{2}} d t \Rightarrow I = \frac{1}{t} + C \Rightarrow I = \frac{1}{\sqrt{1 + \frac{2}{x} cos α + \frac{1}{x^{2}}}} + C \Rightarrow I = \frac{x}{\sqrt{x^{2} + 2 x cos α + 1}} + C \Rightarrow f (x) = x, g (x) = x^{2} + 2 x cos α ∴ g (x) - 2 cos α f (x) = x^{2}$

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If ∫xcosα+1(x2+2xcosα+1)3/2dx=f(x)√g(x)+1+C, then g(x)−2cosαf(x) is (where C is constant of integration)

If $\int \frac{x cos α + 1}{(x^{2} + 2 x cos α + 1)^{3 / 2}} d x = \frac{f (x)}{\sqrt{g (x) + 1}} + C$ , then $g (x) - 2 cos α f (x)$ is
(where $C$ is constant of integration)