Q1. y=sin^-^1(2x\sqrt{1-x^2}) then dy/dx = (Find answer by trigonometric substitution)?
My attempt:
Let x=sinθ
Then the given expression translates to y=2θ = 2sin-1x
So, dy/dx = 2/√1-x2
Let x=cosθ
Then the given expression translates to y=2θ = 2cos-1x
So, dy/dx = -2/√1-x2
Why such discrepancy?
Q2. \int \frac{dx}{(3sinx+4cosx)^2}
11Accordingly, i put tan2x=sec2x-1....So integrand turns out to be......
secx(sec2x-1)2dk, where tanx=k......
finally it becomes ∫(1+k2)1/2.k4dk......
Now?
24I found a similar question searching in some books...So not 100% of my original work
\int \sec^3x\tan^4x dx
Now,\tan^2x=\sec^2x-1
=>\int \sec^3x(\sec^2x-1)^2 dx
=>\int \sec^3x(\sec^4x+1-2\sec^2x) dx
=>\int \sec^7xdx+\int \sec^3xdx-2\int \sec^5x dx
Can you finish it from here?
1sir, if it involves nything out off iit-jee syllabus , then no need.
thnx sir fr informing.
66Are you sure that this is the integral you require. The answer will involve something called the Catalan's constant.
1plzzzz.... solve dis :
∫(0 to 1) 2(tan-1 x)² dx ( tan-1 x = arctanx )
1Ya,got the approach.i'm continuing frm what soumik did.
I = ∫(1+k2)1/2.k 4 dk...... now,put integration by parts.
(1+k2)1/2. k5/5 - ∫(1+k2)-1/2.k . k5/5 dk
I = (1+k2)1/2. k5/5 - 1/5 ∫ [ t6 / √(1+t2)]
I = (1+k2)1/2. k5/5 - 1/5 ∫ [(t2)3+1-1] / √(1+t2)
I = (1+k2)1/2. k5/5 - 1/5 ∫√(1+t2) ( t4+1-t2) [ got by putting formula of a3+b3 ]
I = (1+k2)1/2. k5/5 - 1/5 [ ∫x4 √(1+x2) + ∫√(1+x2) - ∫t2√(1+x2)
now, I = ∫x4 √(1+x2). SO,NOW BRINGING I/5 TO THE L.H.S. ,we cn easily solve the integral.
11Can u plz elaborate the full soln? Actually this is not a very common way, that's the reason i'm so interested....(In case u think the other way....)
19then......
the expression looks like :
∫z√(z2+1) dz
......for this the linear factor (z) is expressed in terms of the derivative of the quadratic factor (z2+1) together with a constant as ,,,,,z=λ{d/dz(z2+1)} + μ...we find the values of λ and μ by equating coefficients of similar terms and then put it the integrand !!!!
..............THAT BECOMES A LOT SIMPLER !!!!!
11For the 2nd one, we can use the universal substitution - that simplifies it a lot, followed by partial fractions....
See the integral turns out to be 2∫1+t24-4t2+6tdx
hopefully the rest can be done....
19the golden rule in such cases : ∫ tanmx secnx dx
(1)if the power of secx is an even positive integer put z= tan x and then integrate
(2)if the power of secx is NOT an even positive integer then see the power of tan x
(a) if the power of tan x is an odd positive integer then put z=secx and then
integrate.
(b)if the power of tan x is an even positive integer put
tan2x=sec2x-1 , then substitute z=tanx and integrate
(3) if the power of tanx is zero and the power of sec x is an odd positive integer>1
then use integration by parts !!!!
so, for the question given here : rule (2)-(b) applies ......rest can be done easily !!!!
11This should do it....
If it is broken in sinx and cosx, it is seen that cosx is raised to an odd power, thus take sinx =t, so it becomes ∫t4(1-t2)-4dt.....Apply By-parts now....
take du=t2
v=-16(1-t2)^3, meaning dv=tdt(1-t2)^3
Now it can be done...(I hope)
106thx bhatt sir and kaymant sir, soumik and debotosh
Q3. \int sec^3xtan^4xdx
66Asish, the expressions for Q2 obtained in #2 and #3 will differ by a constant. Remember the C appearing in the indefinite integration. (I forgot to put that.)
341The discrepancy arises because
1. \sin^{-1} (\sin(x) ) = x is true iff -\frac{\pi}{2} \le x \le \frac{\pi}{2}
2. The expression given by you for \frac{dy}{dx} is not entirely correct
When you put x = sin θ, choose the domain of θ as \left[-\frac{\pi}{2}, \frac{\pi}{2}, \right]
You can verify that
y = 2 \theta when -\frac{\pi}{4} \le \theta \le \frac{\pi}{4}
= \pi - 2 \theta when \frac{\pi}{4} \le \theta \le \frac{\pi}{2}
=-\pi - 2 \theta when -\frac{\pi}{2} \le \theta \le -\frac{\pi}{4}
Now, you can see that
\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}} when x^2 \le \frac{1}{2} and
= -\frac{2}{\sqrt{1-x^2}} when \frac{1}{2} \le x^2 \le 1
You can verify this by directly taking derivative without substitution. Just remember there that \sqrt{(1-2x^2)^2} = 1-2x^2 \ \text{when} \ x^2 \le \frac{1}{2} and
2x^2-1 \ \text{otherwise}
Likewise when you use x = cos θ, there are more cases that need to be taken, so its not worthwhile.
106ok . for Q2. #2 and #3 give two methods. So accordingly they should give the same integral upon expansion. But upon equating the two results that I obtained it doesnt come as a universal relation as 0=0 or like that
I get just two values of x for which the integrals are equal to each other.
66For the second one just note that
3\sin x + 4\cos x = 5\sin(x+\alpha)
where \tan \alpha = \dfrac{4}{3}
As such the integral
I = \int \dfrac{\mathrm dx}{25 \sin^2(x+\alpha)}
=-\dfrac{1}{25}\cot(x+\alpha)
19for (2) :
divide both numerator and denominator by cos2x , replace sec2x(if any) in the denominator by(1 + tan2x)and then put tan x = t,,,,,, so that sec2x dx = dt