Show that $$\int\limits\limits {1-(\tan \theta)^2 \over [1 +(\tan \theta)^2]^2}theta = \int\limits \cos2 \theta d \theta$$

Question

lgbasallote · Answer

no offense..but i really find the inconsistency of your math topics funny =)))) anyway...for this question uhmmm...

anonymous · Answer

dx is on the left dtheta on the right?

lgbasallote · Answer

hmm good point -_-

anonymous · Answer

Yes, it is

lgbasallote · Answer

maybe $1 + \tan^2 \theta = \sec^2 \theta$ will be useful here??

anonymous · Answer

so on the left side, you're integrating wrt theta, and there's no theta... are those theta's supposed t be x's?

shubhamsrg · Answer

miss emma,,that cant be ..please recheck the ques..

shubhamsrg · Answer

i mean both dx and dtheta cant be there

lgbasallote · Answer

i think she already corrected herself that it's $d\theta$ :p

anonymous · Answer

It's meant to come out as theta but it starts of as dx

lgbasallote · Answer

wait..what?

anonymous · Answer

It's not yet integrating.. first it's identities...

anonymous · Answer

That's what the question says

lgbasallote · Answer

so...we need to change it to x...without knowing the relationship of theta to x?

shubhamsrg · Answer

well in that case ,,if we assume ques to be correct,,we should suppose theta is constant with dx ..and then simplify,,which is much easier..
and the solution obtained should be integral of cos2theta dtheta

anonymous · Answer

Yup, but how to get there? I don't know... I'm not good with identities and trig.

shubhamsrg · Answer

well then,,
multiply and divide LHS by $$\cos ^{4}\theta $$

convert to sin and cos terms..

shubhamsrg · Answer

the denominator then becomes 1..

shubhamsrg · Answer

numerator is cos 2(theta) * cos^2 (theta)

shubhamsrg · Answer

which is definately not the integral of cos2(theta)
so please re check the ques,,else i'd assume ques to be incorrect..

anonymous · Answer

OH, yes sorry.. it was first dx, and then I changed it myself to theta, so it should be theta.

shubhamsrg · Answer

i still think its incorrect

anonymous · Answer

It's theta to theta...

anonymous · Answer

That's the question

shubhamsrg · Answer

i re checked my soln and the ques is wrong..

anonymous · Answer

That\s the question here...

anonymous · Answer

It has something to do with dx = sec2^θ dθ

shubhamsrg · Answer

theres a cos^2 (theta) extra coming in the numerator on LHS along with cos 2(theta)

shubhamsrg · Answer

dx=sec^2 theta dtheta ?? whats this?? you derived this ?

anonymous · Answer

In the answer, it says State or imply dx = sec^2θ dθ

shubhamsrg · Answer

yes,,theere we go..in presence os sec^2 theta,,cos^2 theta gets cancelled and we get our answer,,which makes it correct miss watson..

anonymous · Answer

Can you show via latex? It's easier to read and understand for me :D

shubhamsrg · Answer

$$1−(\tanθ)^2[1+(\tanθ)^2]^2 $$ multiply and divide this by (cos(theta))^4
reduce expression into purely sin and cos terms..
the denominator becomes 1 as it is (sin^2(theta) + cos^2(theta))^2 
now num is cos^2(theta) * cos (2theta) 
since you say dx = sec^2(theta),, we are left with only cos(2theta)
hope it was clear now?

lalaly · Answer

is the fraction multiplied by theta? Can u rewrite ur problem ?

lalaly · Answer

When u decide to fix this tag me

experimentx · Answer

this doesn't seem to be a valid identity ... are you sure about this question?? ------------------ http://www.wolframalpha.com/input/?i=integrate+%28%281-tan^2+x%29%2F%281%2Btan^2+x%29^2+from+0+to+pi%2F4 -------------------- http://www.wolframalpha.com/input/?i=integrate+cos+2x+from+0+to+pi%2F4 -------------------- http://www.wolframalpha.com/input/?i=%28%281-tan^2+x%29%2F%281%2Btan^2+x%29^2+%3D+cos+2x

lalaly · Answer

The right identity is$$\large{\frac{1-\tan^2x}{1+\tan^2x}=\cos(2x)}$$

experimentx · Answer

that makes sense!!