Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Integral 1+tan(x)tan(x+a)
I = [1 + tanxtanx+a]dx I = [1 + (sinx/cosx)(sinx+a/cosx+a)] I = [cosx+acosx+sinx+asinx]dx/cosxcosx+a (using , cosacosb+sinasinb = cosa-b) I = (cosa)dx/cosxcosx+a multiply divide by sina I = cota (sina)dx/cosxcosx+a ...............1 sina = sin[(x+a)-x] = sin(x+a)cosx - cos(x+a)sinx ............2 putting 2 in eq 1 I = cota [ tan(x+a) - tanx]dx I = cota [ logsec(x+a) - logsecx] + c I = cota { log[sec(x+a)/secx] } + c this is the required integral approve my ans if u like it
I = [1 + tanxtanx+a]dx
I = [1 + (sinx/cosx)(sinx+a/cosx+a)]
I = [cosx+acosx+sinx+asinx]dx/cosxcosx+a (using , cosacosb+sinasinb = cosa-b)
I = (cosa)dx/cosxcosx+a
multiply divide by sina
I = cota (sina)dx/cosxcosx+a ...............1
sina = sin[(x+a)-x] = sin(x+a)cosx - cos(x+a)sinx ............2
putting 2 in eq 1
I = cota [ tan(x+a) - tanx]dx
I = cota [ logsec(x+a) - logsecx] + c
I = cota { log[sec(x+a)/secx] } + c
this is the required integral
approve my ans if u like it
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -