Doubts

I 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?

sir, if it involves nything out off iit-jee syllabus , then no need.

thnx sir fr informing.

plzzzz.... solve dis :
∫(0 to 1) 2(tan^-1 x)² dx ( tan^-1 x = arctanx )

Ya,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) - ∫t²âˆš(1+x²)

now, I = ∫x⁴ √(1+x²). SO,NOW BRINGING I/5 TO THE L.H.S. ,we cn easily solve the integral.

anyone got the answer of #8 yet?

Can 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....)

then......
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 !!!!!

For 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....

the golden rule in such cases : ∫ tan^mx 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 !!!!

thx bhatt sir and kaymant sir, soumik and debotosh

Q3. \int sec^3xtan^4xdx

Asish, 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.)

The 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.

ok . 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.