Let x2 - n - - 1- n - N- Then- the value of - x -2 sin x2 - 1 - sin 2 x2 - 1-2 sin x2 - 1 - sin 2 x2 - 1 dx is equal to

We have, ∫ x √(2 sin (x^2 + 1) sin 2 (x^2 + 1)/2 sin (x^2 + 1) + sin 2 (x^2 + 1))dx = ∫ x √(2 sin (x^2 + 1) 2 sin (x^2 + 1) cos (x^2 + 1)/2 sin (x^2 + 1 ...

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Standard XII

Mathematics

Standard Formulae - 2

Let x2 ≠ n π ...

Question

Let $x^{2} \neq n π - 1, n ϵ N$ . Then, the value of $\int x \sqrt{\frac{2 s i n (x^{2} + 1) - s i n 2 (x^{2} + 1)}{2 s i n (x^{2} + 1) + s i n 2 (x^{2} + 1)}} d x$ is equal to

l o g | \frac{1}{2} s e c (x^{2} + 1) | + C

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l o g | s e c (\frac{x^{2} + 1}{2}) | + C

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\frac{1}{2} l o g | s e c (x^{2} + 1) | + C

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None of these

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Solution

The correct option is B $l o g | s e c (\frac{x^{2} + 1}{2}) | + C$
We have, $\int x \sqrt{\frac{2 s i n (x^{2} + 1) - s i n 2 (x^{2} + 1)}{2 s i n (x^{2} + 1) + s i n 2 (x^{2} + 1)}} d x = \int x \sqrt{\frac{2 s i n (x^{2} + 1) - 2 s i n (x^{2} + 1) c o s (x^{2} + 1)}{2 s i n (x^{2} + 1) + 2 s i n (x^{2} + 1) c o s (x^{2} + 1)}} d x = \int x \sqrt{\frac{1 - c o s (x^{2} + 1)}{1 + c o s (x^{2} + 1)}} d x = \int x t a n (\frac{x^{2} + 1}{2}) d x = \int t a n (\frac{x^{2} + 1}{2}) d (\frac{x^{2} + 1}{2}) = l o g | s e c (\frac{x^{2} + 1}{2}) | + C$

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Similar questions

Q. For

x^{2} \neq n π + 1, n ϵ N

(the set of natural numbers), the integral

\int x \sqrt{\frac{2 sin (x^{2} - 1) - sin 2 (x^{2} - 1)}{2 sin (x^{2} - 1) + sin 2 (x^{2} - 1)}} d x

is equal to
(where

c

is a constant of integration).

\int x . \sqrt{[\frac{2 sin (x^{2} - 1) - sin 2 (x^{2} - 1)}{2 sin (x^{2} - 1) + sin 2 (x^{2} - 1)}]} d x

where

x^{2} - 1 \neq n π

Q. Let

x^{2} \neq n π - 1, n ϵ N

. Then, the value of

\int x \sqrt{\frac{2 s i n (x^{2} + 1) - s i n 2 (x^{2} + 1)}{2 s i n (x^{2} + 1) + s i n 2 (x^{2} + 1)}} d x

is equal to

Q. For

x^{2} \neq n π + 1, n \in N

(the set of natural numbers), the integral

\int x \sqrt{\frac{2 sin (x^{2} - 1) - sin 2 (x^{2} - 1)}{2 sin (x^{2} - 1) + sin 2 (x^{2} - 1)}} d x

is equal to :
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Q. Solve equation for general solution

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Let x2≠nπ−1,nϵN. Then, the value of ∫x√2sin(x2+1)−sin2(x2+1)2sin(x2+1)+sin2(x2+1)dx is equal to

The correct option is B log|sec(x2+12)|+CWe have, ∫x√2sin(x2+1)−sin2(x2+1)2sin(x2+1)+sin2(x2+1)dx =∫x√2sin(x2+1)−2sin(x2+1)cos(x2+1)2sin(x2+1)+2sin(x2+1)cos(x2+1)dx =∫x√1−cos(x2+1)1+cos(x2+1)dx =∫xtan(x2+12)dx =∫tan(x2+12)d(x2+12) =log|sec(x2+12)|+C

Let $x^{2} \neq n π - 1, n ϵ N$ . Then, the value of $\int x \sqrt{\frac{2 s i n (x^{2} + 1) - s i n 2 (x^{2} + 1)}{2 s i n (x^{2} + 1) + s i n 2 (x^{2} + 1)}} d x$ is equal to