wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Evaluate:

1)x3+3x+4xdx

2)x(3x2+2x+3)dx

3)(2x23sinx+5x)dx

4) sec2xcsc2xdx



Open in App
Solution

(I)
x3+3x+4xdx

(x31/2+3x11/2+4x1/2)dx

x5/2+3x1/2+4x1/2

x7/27/2+x3/23/2+8x1/2+c

(II)
x(3x2+2x+3)dx

(3x5/2+2x3/2+3x1/2)dx

3x7/27/2+2x5/25/2+3x3/23/2+c

(III)
(2x23sinx+5x)dx

2x33+3cosx+5x3/23/2+c

(IV)
sec2xcosec2xdx

=tan2xdx

(sec2x+1)dx

tanxx+c

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Standard Formulae 1
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon