Indefinte integration reloaded

Loser!

1) $\int \frac{(x-1)e^{x}}{(x+1)^{3}}$

2) $\int log(\sqrt{1-x}+\sqrt{1+x})dx$

3) $\int_{0}^{a}\sqrt{a^{2}-x^{2}}(cos^{-1}\frac{x}{a})^{2}dx$

4) $\int_{0}^{1}cot^{-1}(1+x^{2}-x)$

akki ~~ unlucky forever ~~

akki ~~ unlucky forever ~~

akki ~~ unlucky forever ~~

$put\ x=acos\theta\\ \\ I=a^2\ \int_{0}^{\frac{\pi}{2}}\theta^2sin^2\theta\ d\theta\\ \\ or\ I = a^2\ \int_{0}^{\frac{\pi}{2}}cos^2\theta\ (\theta^2+\frac{\pi^2}{4}-\theta\pi)d\theta\\ \\adding\ both\\ \\ I=\frac{a^2}{2}\int_{0}^{\frac{\pi}{2}}(\theta^2+\frac{\pi^2}{4}cos^2\theta-\pi\theta\ cos^2\theta)d\theta$

now,all 3 integrals are easily seperately integrable . . .

akki ~~ unlucky forever ~~

$I=\int_{0}^{1}tan^{-1}\frac{1}{(1-x+x^2)}dx\\ \\ I=\int_{0}^{1}tan^{-1}(\frac{(1-x )+ x}{1-x(1-x)})dx\\ \\ I=\int_{0}^{1}tan^{-1}(1-x)\ dx+\int_{0}^{1}tan^{-1}x\ dx\\ \\ applying\ integration\ b\ parts\ to\ each\\ \\ I=\int_{0}^{1}\frac{1-x}{1+x^2}dx+\frac{\pi}{4}-\int_{0}^{1}\frac{x\ dx}{1+x^2}\\ \\ I=\frac{\pi}{2}-\int_{0}^{1}\frac{2x\ dx}{1+x^2}\\ \\ I=\frac{\pi}{2}-ln2$

.

kabi

2. using integration by parts

I = x log (Sqrt(1-x)+sqrt(1+x)) - integration ( ( 1-Sqrt(1-x²)) /( -2Sqrt(1-x²)) dx

= x log (Sq.(1-x)+Sq (1+x)) - 1/2 ( integration ( 1 - (1/Sqrt(1-x² )) dx )

= x log (Sq.(1-x)+Sq (1+x))- x/2 +sin^-1(x)/2 +constant

thank u

Loser!

Thanx for the replies! I have few more doubts..plz try these

5) $\int cos 2\theta log\left ( \frac{cos\theta+sin\theta }{cos\theta -sin\theta } \right )d\theta$

6) $ST \int e^{tan^{-1}x}\left ( \frac{1+x+x^2}{1+x^{2}} \right )dx=xe^{tan^{-1}x}+c$

7) $\int xe^{x}cosx dx$

8) $\int log(1+cosx)-xtan(\frac{x}{2})$

taran

first can be done by parts

= cos2theta * log [tan(pie/4 +theta)]

=1/2*log [tan(pie/4 +theta)]* sin2theta - integral of 1/ tan(pie/4 +theta) * sec^2 (pie/4 +theta ) *sin2theta /2

after simplification of second integrand we get

=1/2*log [tan(pie/4 +theta)]* sin2theta - inetgral of 2tan2theta

{ bcoz 1/ tan(pie/4 +theta) * sec^2 (pie/4 +theta ) = 1/ sin(pie/4 +theta)cos(pie/4 +theta) = 2/sin2(pie/4 +theta) = 2/ sin(pie/2 +2theta) = 2/ cos2theta which is in multiplication with sin2theta so becomes tan2theta}

=1/2*log [tan(pie/4 +theta)]* sin2theta - log mod sec2theta

taran

6>

= e^tan inv x { 1 + x/(1+x^2 )}dx

substitute tan invx x =y

= e^y { 1+ tany /sec^2 y } sec^2 y dy

= e^y { sec^2 y +tany }dy

=e^y tany (integration by parts )

now resubsitute

= x e^tan inv x

taran

7> this one's is easy

just take x as first function and e^x cosx as second function

then integrate

taran

8>

edited

Akshay

8) The reply given above is wrong as u cant solve further

Instead use by parts directly keeping the second integral as it is.

x.log[1 + cosx] - ∫x.(-sinx)/(1 + cosx)dx - ∫xtan(x/2)dx

x.log[1 + cosx] + ∫x. [ 2 sin(x/2)cos(x/2)]/[2cos^2 (x/2)]dx - ∫xtan(x/2)dx

xlog[1 + cosx] +∫xtan(x/2)dx - ∫xtan(x/2)dx

xlog[1 + cos(x/2)] + c

taran

oops sorry ...thnks for pointing it out

The wicked warrior ''Anurag the devil''

edited

The wicked warrior ''Anurag the devil''

edited

The wicked warrior ''Anurag the devil''

jus take x=cos2y and then use by parts twice

Akshay

7) Integrating by parts

Taking x as first function and e^x cosx as second function

x ∫cosx e^xdx - ∫ ∫cosx e^xdxdx -----(i)

Now solving ∫e^x(cosx)dx

Let I = ∫e^x(cosx)dx

solving by parts let cosx be first function

I = e^x(cosx) - ∫e^x(-sinx)dx

I = e^x(cosx) + ∫e^x(sinx)dx

Again byparts sinx is first function

I = e^x(cosx) + e^x(sinx) - ∫e^x(cosx)dx

I = e^x(cosx +sinx) - I

2I = e^x(cosx +sinx)

I = e^x(cosx +sinx)/2

Now substituting in integral (i)

xe^x(cosx + sinx)/2 -∫e^x(cosx +sinx)/2dx

xe^x(cosx +sinx)/2 - e^x(sinx)/2 + c [ By using the formula ∫e^x{f(x) + f'(x)}dx = e^x(f(x))]

The wicked warrior ''Anurag the devil''

in that fourth question first he has converted the inverse cotangent function to inverse tangent function,,,,,,then the next step u hav already understood then he has used tan inverse x + tan inverse y=tan inverse { (x+y)/1-xy}.........then he has applied integration by parts this question is also their in a das gupta and arihant u can see it over there also the method is same bcoz i've never seen any other method for this question as i've already solved this question last month..............

νιѕнαℓ

Loser!

9)If $I_{n}=\int_{0}^{1}x^{n}tan^{-1}xdx$ ,then show that $(n+1)I_{n}+(n-1)I_{n-2}=\frac{\pi }{2}-\frac{1}{n}$

10)Prove that $\int_{0}^{\frac{\pi }{4}}xcosxcos3xdx=\frac{\pi -3}{16}$

11) $\frac{x-sinx}{1-cosx}$

12)show that $\int_{t}^{1}\frac{dx}{1+x^{2}}=\int_{1}^{\frac{1}{t}}\frac{dx}{1+x^{2}}$

Ask & Discuss Questions with Community & Experts