bitsat 2009

JEE Advanced 2015 Coaching › Calculus questions › bitsat 2009

solve question of bitsat i m not able to provide option

#Mathematics #Calculus

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8 Answers

1

prateek punj ·2009-05-16 02:28:54

1) take x=tanÎ¸

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1

rahul wadhwani ·2009-05-16 02:31:56

why r u not using biparts taking u= 1 and v= tan-1 x

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1

prateek punj ·2009-05-16 02:33:10

yeah u can do it with that too...

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rahul wadhwani ·2009-05-16 02:34:31

ok. i know more quest. but i have less time ok next time bye

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1

prateek punj ·2009-05-16 02:34:35

4) rationalize...

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11

Mani Pal Singh ·2009-05-17 09:32:16

1)
use by parts
\int_{0}^{1}{tan^{-1}x}=x.tan^{-1}x|_{0}^{1}-\int_{0}^{1}{\frac{x}{1+x^{2}}}

i think u r facing trouble to solve it
multiply and divide it by 2
then take
1+x²=t
2xdx=dt

so u get
ln(1+x²)|₀¹

=pi/4-ln2

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11

Mani Pal Singh ·2009-05-17 09:33:54

2)
multiply and divide the eq by 2
u get
1/2[âˆ«cos5x + âˆ«cosx]
hence
1/2[sin5x/5+sinx]

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11

Mani Pal Singh ·2009-05-17 09:37:37

Q 3 and Q 4 are same .Just take a common in 4)
for Q3)
u see that
âˆ«1-cosx/sin²x dx
âˆ«cosec²x-âˆ«cotxcosecxdx
=-cotx+cosecx

and 4 the 4) u will get 1/a(-cotx+cosecx)

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Your Answer

Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

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° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω