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Question Number 65011 by AnjanDey last updated on 24/Jul/19

1 . (i) Evaluate : \int \frac{1}{\sin x - \cos x + \sqrt{2}} d x

(ii) Evaluate : \int 2^{2^{2^{x}}} 2^{2^{x}} 2^{x} d x

(iii) Evaluate : \int \frac{\cos^{3} x}{\sin^{2} x + \sin x} d x

2 . cosec [\tan^{- 1} {\cos (\cot^{- 1} (\sec (\sin^{- 1} a)))}] = What ?

3 . Prove that, \sin [\cot^{- 1} {\cos (\tan^{- 1} x)}] = \sqrt{\frac{x^{2} + 1}{x^{2} + 2}}

4 . Mention Order and Degree and state also if it is linear or non - l inear .

y + \frac{d^{2} y}{{d x}^{2}} = \frac{19}{25} \int y^{2} d x

Commented by mathmax by abdo last updated on 24/Jul/19

1) l e t I = \int \frac{d x}{s i n x - c o s x + \sqrt{2}} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

I = \int \frac{1}{\frac{2 t}{1 + t^{2}} - \frac{1 - t^{2}}{1 + t^{2}} + \sqrt{2}} \frac{2 d t}{1 + t^{2}} = \int \frac{2 d t}{2 t - 1 + t^{2} + \sqrt{2} + \sqrt{2} t^{2}}

= \int \frac{2 d t}{(1 + \sqrt{2}) t^{2} + 2 t + \sqrt{2} - 1}

Δ^{^{'}} = 1 - (2 - 1) = 0 \Rightarrow o n e r o o t s x_{0} = \frac{- b^{^{'}}}{a} = \frac{- 1}{1 + \sqrt{2}} \Rightarrow

I = \frac{1}{1 + \sqrt{2}} \int \frac{2 d t}{{(t + \frac{1}{1 + \sqrt{2}})}^{2}} + C = \frac{- 2}{(1 + \sqrt{2}) (t + \frac{1}{1 + \sqrt{2}})} + C = \frac{- 2}{(1 + \sqrt{2}) t + 1} + C

= \frac{- 2}{1 + (1 + \sqrt{2}) t a n (\frac{x}{2})} + C .

Answered by Tanmay chaudhury last updated on 24/Jul/19

1) \frac{1}{\sqrt{2}} \int \frac{d x}{1 + s i n (x - \frac{π}{4})}

\frac{1}{\sqrt{2}} \int \frac{1 - s i n (x - \frac{π}{4})}{{c o s}^{2} (x - \frac{π}{4})} d x

\frac{1}{\sqrt{2}} \int {s e c}^{2} (x - \frac{π}{4}) - t a n (x - \frac{π}{4}) s e c (x - \frac{π}{4}) d x

\frac{1}{\sqrt{2}} [t a n (x - \frac{π}{4}) - s e c (x - \frac{π}{4})] + c

Commented by AnjanDey last updated on 24/Jul/19

A l t e r :

I = \int \frac{1}{\sin x - \cos x + \sqrt{2}} d x

H e r e, \sin x - \cos x + \sqrt{2} = \sqrt{2} - \sqrt{2} (\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x)

= \sqrt{2} {1 - \cos (\frac{π}{4} + x)}

= 2 \sqrt{2} \sin^{2} (\frac{π}{2} + 2 x)

= 2 \sqrt{2} \cos^{2} 2 x

∴ I = \int \frac{1}{2 \sqrt{2} \cos^{2} 2 x} d x

= \frac{1}{2 \sqrt{2}} \int \sec^{2} 2 x d x

= \frac{1}{2 \sqrt{2}} \cdot \frac{\tan 2 x}{2} + C (W h e r e C = c o n s t a n t o f i n t e g r a t i o n)

= \frac{1}{4 \sqrt{2}} \tan 2 x + C

I t m a y b e e a s y m e t h o d \dots

Answered by Tanmay chaudhury last updated on 24/Jul/19

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

\int \frac{(1 - s i n x) c o s x d x}{s i n x}

\int \frac{(1 - t)}{t} d t [t = s i n x \frac{d t}{d x} = c o s x]

\int \frac{d t}{t} - \int d t

= l n t - t + c

= l n (s i n x) - s i n x + c

Answered by Tanmay chaudhury last updated on 24/Jul/19

3) {t a n}^{- 1} x = θ t a n θ = x

s i n [{c o t}^{- 1} {c o s ({t a n}^{- 1} x)}]

= s i n [{c o t}^{- 1} {c o s θ}]

l e t {c o t}^{- 1} (c o s θ) = α

c o t α = c o s θ = \frac{1}{\sqrt{1 + x^{2}}} [s i n c e t a n θ = x c o s θ = \frac{1}{\sqrt{1 + x^{2}}}]

s i n (α) = \frac{1}{c o s e c α}

s i n α = \frac{1}{\sqrt{1 + {c o t}^{2} α}} = \frac{1}{\sqrt{1 + \frac{1}{1 + x^{2}}}}

s i n α = \frac{\sqrt{1 + x^{2}}}{\sqrt{2 + x^{2}}}

Answered by Tanmay chaudhury last updated on 24/Jul/19

2) {s i n}^{- 1} a = θ

s e c θ = \frac{1}{c o s θ} = \frac{1}{\sqrt{1 - {s i n}^{2} θ}} = \frac{1}{\sqrt{1 - a^{2}}}

c o s e c [{t a n}^{- 1} {c o s ({c o t}^{- 1} (s e c ({s i n}^{- 1} a)))}]

= c o s e c [{t a n}^{- 1} {c o s ({c o t}^{- 1} (\frac{1}{\sqrt{1 - a^{2}}}))}]

{c o t}^{- 1} (\frac{1}{\sqrt{1 - a^{2}}}) = α

c o t α = \frac{1}{\sqrt{1 - a^{2}}} s o c o s α = \frac{1}{\sqrt{2 - a^{2}}}

c o s e c [{t a n}^{- 1} {\frac{1}{\sqrt{2 - a^{2}}}}]

t a n β = \frac{1}{\sqrt{2 - a^{2}}} \to s i n β = \frac{1}{\sqrt{3 - a^{2}}}

c o s e c [{t a n}^{- 1} {t a n β}]

c o s e c β

= \sqrt{3 - a^{2}}

Answered by Tanmay chaudhury last updated on 24/Jul/19

\int 2^{2^{2^{x}}} 2^{2^{x}} 2^{x} d x

t = 2^{x}

\frac{d t}{d x} = 2^{x} l n 2

\frac{1}{l n 2} \int 2^{2^{t}} 2^{t} d t

k = 2^{t}

\frac{d k}{d t} = 2^{t} l n 2

\frac{1}{l n 2} \times \int 2^{k} \times \frac{d k}{l n 2}

= \frac{1}{{(l n 2)}^{2}} \int 2^{k} d k

= \frac{1}{{(l n 2)}^{3}} \times 2^{k} + c

= \frac{1}{{(l n 2)}^{3}} \times 2^{2^{t}} + c

= \frac{1}{{(l n 2)}^{3}} \times 2^{2^{2^{x}}} + c