Calculus - continue d thread

4

UTTARA ·Mar 24 '10 at 4:12

U can use the formula directly for sec^n (2x) or this may work out

Take 2x = t

Using IBP

sect âˆ«(sec^2 (t) dt = sec t tan t

âˆ«sec t tan t (tan t ) dt = âˆ« p² - 1 dp

Put sec t = p

sec t tan t dt = dp

Now I = 1/2(sec t tan t - sec³t - sec t )

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4

UTTARA ·Mar 24 '10 at 4:19

2 ) I = âˆ« (pi-x) /(a² - cos² x ) dx

2I = pi âˆ«1/(a² - cos² x ) dx

Put tanx = p

sec² x dx = dp

âˆ« 1/[(a² - 1) + a² p²] dp

= 1/(a² - 1) âˆ« 1/[1+kp²] dp ---> I took k = const = a²/(a²-1)

= 1/(a² - 1) tan^-1 (kp) + c

I = (pi/2)[1/(a² - 1)] tan^-1 [(a²)/(a²-1)(tanx)] ] (0 to pi)

= 0

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1

" ____________ ·Mar 24 '10 at 5:46

FOR SECOND QUES

a is supposed to b greater than 1( a>1)

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1

" ____________ ·Mar 25 '10 at 2:13

any one tryin for ques --2

wil post answers if neone doesnt post answer widin today

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" ____________ ·Mar 25 '10 at 2:17

ans is

Ï€²2a.√a²-1

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29

govind ·Mar 25 '10 at 12:58

SR have asked me many times to post some integrals..so SR i am posting them....

1.. $\int 1 2 n + 1 (x - 1) (x - 2) (x - 3) . . . . (x - (2 n + 1)) d x$
where n is a natural number..

2... $\int 0 i n f i n i t y x 1 s i n (x - x 1 ) d x$

3.. $\int s i n - 1 \sqrt 4 x 2 + 8 x + 13 2 x + 2 d x$

4.. $\int 0 π / 4 e s i n x (x c o s x - s e c x t a n x) d x$

5.. $\int 0 i n f i n i t y 1 + x 2 1 l n (x + x 1 ) d x$

6.. $\int 0 i n f i n i t y (1 + x 2 ) 2 x l n x d x$

For Practise

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1

" ____________ ·Mar 26 '10 at 1:22

1. ans is 0

( odd function )

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1

" ____________ ·Mar 26 '10 at 3:52

3. iit-jee question ( which all of u must have solved )

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b_k_dubey ·Mar 26 '10 at 7:45

$\int e s i n x x c o s x d x - \int e s i n x s e c x t a n x d x$

$= x \int e s i n x c o s x d x - \int ( d x d x \int e s i n x c o s x d x) d x - \int e s i n x s e c x t a n x d x$

$= x e s i n x - \int e s i n x d x - \int e s i n x s e c x t a n x d x$

$= x e s i n x - \int e s i n x s e c x c o s x d x - \int e s i n x s e c x t a n x d x$

$x e s i n x - \int s e c x d (e s i n x ) - \int e s i n x d (s e c x)$

$= x e s i n x - \int d (e s i n x s e c x)$

$= x e s i n x - e s i n x s e c x$

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1

Manmay kumar Mohanty ·Mar 26 '10 at 21:39

6)
Integral is : $\int 0 \infty (1 + x 2 ) 2 x l n x d x$
first i solved it indefinite terms so i took $\int (1 + x 2 ) 2 x l n x d x$
now put 1+x² = t $\Rightarrow 2xdx= dt$
now tntegral can be written as $ 2 1 \int (1 + x 2 ) 2 2 x l n x d x$ $\Rightarrow \frac{1}{2}\int \frac{ln(\sqrt{t-1})}{t^{2}}dt$

hence considering $ t 2 1 $ as scond and $l n (\sqrt t - 1 )$ as first function we have integral as
$ 2 1 \int t 2 1 . l n (\sqrt t - 1 ) d t$
now
$\Rightarrow \frac{1}{2}\left[ln(\sqrt{t-1}) \int \frac{1}{t^{2}}dt+\int \frac{1}{\sqrt{t-1}}.\frac{1}{2\sqrt{t-1}}.\frac{1}{t}dt\right]$

$ 2 1 [l n (\sqrt t - 1 ) ( t - 1 ) + 2 1 \int (t - 1) t 1 d t]$
and this can be finished off now

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1

" ____________ ·Mar 29 '10 at 3:36

2 .PUT 1/x=t

5th one one

by parts does the question

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