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Integral Calculus

Blazing goIITian

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16 Jan 2010 10:43:25 IST

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plz someone solve this one i'm able to integrate it but the primitive is not defined at 0........int

None

plz someone solve this one i'm able to integrate it but the primitive is not defined at 0........intergattion (from 0 to pi /2) 8+7 cos x.dx/(8cosx + 7 )^2...

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DEVENDRA MEHENDALE

Blazing goIITian

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18 Jan 2010 11:17:18 IST

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Use substitution in such a way that the primitive is continious at 0.

The wicked warrior ''Anurag the devil''

Blazing goIITian

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18 Jan 2010 15:05:56 IST

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well i've not used substitution i've used the following method.................the numerator we can express as (7cot x + 8 cosec x)sinx and now we can apply by parts by taking (7cot x + 8 cosec x) as first function and remaing as second function.....

TEJASWI V SHETTY.

Hot goIITian

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20 Jan 2010 19:03:15 IST

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u can do dat only wen the two functions are distinct like if one is trigo then the other might b an exponential or an algebriac function. here simply sep. the numerator in two parts and solve.

The wicked warrior ''Anurag the devil''

Blazing goIITian

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20 Jan 2010 22:49:30 IST

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hello i tried that querstion in u'r manner but that will be a very long process and u may not get the ans in that way so i've used the above mentioned method tell me how to evaluate the limits i.e the primitive at 0

Yagyadutt Mishra

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21 Jan 2010 19:09:07 IST

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Answer is sinx /(8cosx+7)

by applying Limit we get 1/7

The wicked warrior ''Anurag the devil''

Blazing goIITian

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22 Jan 2010 20:05:14 IST

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bro plz tell me d method u see i've mentioned my method above tell me if u know any other method.....

shashank agarwal

Blazing goIITian

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22 Jan 2010 20:12:01 IST

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Use substitution in such a way that the primitive is continious at 0.

Answer is sinx /(8cosx+7)

by applying Limit we get 1/7

ITZ VERY LENGTHY , BUT TRY TO SOLVE IT WITH SPEED , ITZ EASY

jee-do or die

Blazing goIITian

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23 Jan 2010 20:11:51 IST

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an easy approach is given in arihant calculus book for these kind of problems

jee-do or die

Blazing goIITian

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23 Jan 2010 20:27:45 IST

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$\int \frac{8( cos^2x + sin ^2x )+ 7 cosx }{( 8 cosx + 7 )^2 }$

$\int \frac{cosx ( 8cosx+ 7 ) + 8 sin^2 x}{ ( 8 cos x + 7 )^2 } dx$

$\int \frac{ ( 7 + 8 cosx) d( sin x) - sin x d( 7 + 8 cos x)}{ 8cosx + 7 }^2 dx$

$\int d\left ( \frac{sinx }{8cosx + 7 } \right )$

$\frac{sinx }{8cosx + 7 }$

The wicked warrior ''Anurag the devil''

Blazing goIITian

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24 Jan 2010 18:36:06 IST

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ok thanx to everyone bt can anyone tell me that what was the problem with my method

jee-do or die

Blazing goIITian

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24 Jan 2010 21:03:55 IST

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i think u r a subscriber of brilliant's correspondence course and they solve very similar to u

there is no problem with ur method by ur method the final answer comes out to be

$\frac{8 cosec x + 7 cot x }{7( 8 + 7 cos x)} - \frac{1}{7} cot x + c$

applying limits u get

$\frac{ 8+ 7 (0 )}{7 ( 8 + 0 )} - \frac{1}{7} ( 0 ) - 0$

HENCE ANS IS $\frac{1}{7}$

I DONT FEEL ANY WRONG WITH UR METHOD

The wicked warrior ''Anurag the devil''

Blazing goIITian

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2 Feb 2010 18:23:35 IST

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yes u r absolutely correct

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