Integration sikhao!!!

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 ...

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]

putting r/n---->t, dr/n----dt

It became
\int_{0}^{2}{\frac{t\; dt}{\sqrt{1+t^2}}}

Correct till now??

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

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

yaar dekh iss mein
put r/n------>t
then 1/n-------->dt

this could be represented in the form of integral
lower limit and upper limit can be found out in the following method

For lower limit
sigma ke nicche waali term check kar
aggar woh constant hai to u put lower limit as 0

For upper limit
check the coefficient of n in the upper power of sigma
the coefficient must be written in the upper limit

so

noe the ques becomes

∫₀¹x⁹⁹dx

and hence 1/100
[1]

Pls post complete solution with easy-to-understand reasoning

Exclude terminologies as much as possible

OK then....

\lim_{n\rightarrow \infty}\frac{1}{n^{100}}\sum_{r=1}^{n}{r^{99}}

(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]

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

subj. 3)

∫dx/sinx(3+2cosx)

= ∫sinxdx/sin²x(3+2cosx)

= ∫sinxdx/(1-cos²x)(3+2cosx)

now put cosx=t...

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]

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 ]