Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Solve pls ∫tan -1 (1-x+x 2 ) dx Limit if from 0 to 1.

Solve pls
∫tan-1(1-x+x2) dx
Limit if from 0 to 1.

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:
Hello Student,
Please find answer to your question below

I = \int_{0}^{1}tan^{-1}(1-x+x^2)dx
I = \int_{0}^{1}cot^{-1}(\frac{1}{1-x+x^2})dx
I = \int_{0}^{1}cot^{-1}(\frac{1}{1-x(1-x)})dx
I = \int_{0}^{1}(\frac{\pi }{2}-tan^{-1}(\frac{1}{1-x(1-x)}))dx
I = \int_{0}^{1}(\frac{\pi }{2}-(tan^{-1}x-tan^{-1}(1-x))dx
I = \int_{0}^{1}\frac{\pi }{2}dx-\int_{0}^{1}tan^{-1}xdx+\int_{0}^{1}tan^{-1}(1-x)dx
I = \int_{0}^{1}\frac{\pi }{2}dx-\int_{0}^{1}tan^{-1}xdx+\int_{0}^{1}tan^{-1}(1-(1-x))dx
I = \int_{0}^{1}\frac{\pi }{2}dx-\int_{0}^{1}tan^{-1}xdx+\int_{0}^{1}tan^{-1}xdx
I = \frac{\pi }{2}

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free