Integration sikhao!!!

Strider koi question pooch
usse samajhenge[1]

6)

(5-4x-x²)^5/2 = (5+x)^5/2(1-x)^5/2

now write numerator as : 1/6[(5+x) + (1-x)]

then the integration bcmes:

1/6 [ ∫(5+x)/(5+x)^5/2(1-x)^5/2 dx + ∫(1-x)/(5+x)^5/2(1-x)^5/2 dx. ]

=1/6[ ∫dx/(5+x)^3/2(1-x)^5/2] + ∫dx/(5+x)^5/2(1-x)^3/2 ]

now apply the trick of maverick...

remeber that is a standard trick ...

8) put x =tanθ
and dx =sec²Î¸dθ

and put y =sin^-1p
and dy =1/√1-p²dp

naw change the limits and the question is done[1]

PLEASE COMPARE THE QUESTIONS WITH JEE LEVEL OF TOUGHNESS AND THEN SOLVE:

#5)

(Subjective again)

\int \frac{(x-1)\; dx}{(x+1)\sqrt{x^3+x^2+x}}

#6)

\int \frac{dx}{\left(5-4x-x^2 \right)^{\frac{5}{2}}}

#7)

\int \frac{dx}{(x-1)^\frac{3}{4}(x+2)^\frac{5}{4}}}

#8)

Prove that

\int_{0}^{1}{\frac{tan^{-1}x\; dx}{x}}=\int_{0}^{\frac{\pi }{2}}{\frac{y\; dy}{siny}}

that'll become......

\left[\sqrt{1+t^2} \right]^{2}_{0}

Hence....(√5-1)??

SOLVE THE FULL QUESTION
UWIONT GET ANYTHING FOR HALF SOLUTION[2]

DO IT THEN WE WILL CHECK[1]

\lim_{n->infinity} 1/n\sum_{r=1}^{2n}{}r/\sqrt{n^{2}+r^{2}}

THIS ONE IS FOR ANIRUDH ONLY!!!!!!!!

yes i may say
in most of the other cases u have to struggle hard to find it
but it of premeir importance
chal i give u a ques 4 practice

What should we look to do in such questions??

Always substitute r/n--->t??

Pls post complete solution with easy-to-understand reasoning

Exclude terminologies as much as possible

(II)

Subjective type

(1)\int sin^3x\; cos(\frac{x}{2})dx

(2)\int sin^{-1}\left(\sqrt{\frac{x}{a+x}} \right)dx

(3)\int \frac{dx}{sinx(3+2cosx)}

(4)\int \frac{sinx\; dx}{sin3x}

Pls give me the algorithm for converting summation to a definite integral...

Also pls illustrate with an example!

[11] [11] [11] [11] [11] [11] [11]

2ND
UR ANSWER WAS SUPERIOR TO MINE JI(LADKION SE NO BEHASS )

@manipal.... afetr where i left it .. in 2nd questions...

its jus one step...

becoz tanθsec²Î¸dθ taken as second function... a child can integrate..

anyways... u have given another method... dat way its nice :)

arey no no... all those partial fractions will become by parts !! [3] sorry..

partial fractions??

GUESSED SO!![3] [3]

Q2
I THINK IT WOULD BE BETTER TO PUT
x+a=t²
dx=2tdt

upper niche waala t kat jaaega
so the ques becomes
∫sin^-1√t²-adt
this could be found out by PARTS (EDITED)[1]

4) ∫sinx/sin3x = ∫sinx/sinx[3-sin2x] = ∫1/[2+cos²x]

=∫sec²xdx/[2sec²x+1]

=∫sec²xdx/[3+2tan²x]

tanx =t le lena ......

then it becomes simple quadratic..

subjective:

2) sin^-1√x/(x+a)

take x=atan²Î¸ => dx = 2atanθsec²Î¸dθ

∫sin^-1√x/(x+a)

=∫sin^-1(sinθ) .2atanθsec²Î¸dθ

= 2a∫θtanθsec²Î¸dθ

now do partial fractions taking θ as first function and tanθsec²Î¸ as second function.

final ans : a[((x/a) +1)tan^-1√x/a - √x/a ]

subjctive tpe
3rd
tan(x/2)=t
proceed aajayega

ni aaya toh kck knck