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

If I=dx2sinx+secx4. Then find the value of I.


Open in App
Solution

Step 1: Simplify the given integral

The given integral can be simplified as,

I=dx2sinx+secx4

I=1cos4x1cos4x×12sinx+secx4dx

I=1cos4x2sinx+secx4cos4xdx

I=sec4x2sinx+secxcosx4dx [secx=1cosx]

I=sec4x2sinxcosx+secxcosx4dx

I=sec4x2tanx+secx·secx4dx [tanx=sinxcosx]

I=sec4x2tanx+sec2x4dx

I=sec4x2tanx+1+tan2x4dx [sec2x=1+tan2x]

I=sec4x1+tanx24dx

I=sec2x·sec2x1+tanx8dx …(i)

Step 2: Assume and deduce values

Now, let, 1+tanx=t …(ii)

tanx=t-1

tan2x=t-12 [Squaring on both sides]

sec2x-1=t-12 [sec2x=1+tan2x]

sec2x=t-12+1 …(iii)

Differentiating equation (ii) with respect to x,

ddx1+ddxtanx=dtdx

0+1cos2x=dtdx [ddxC=0,ddxtanx=1cos2x]

sec2x=dtdx [secx=1cosx]

sec2xdx=dt …(iv)

Step 3: Substitute values and integrate

Now, substituting the values obtained in equations (ii), (iii) and (iv) into the equation (i), we get,

I=t-12+1t8dt

I=t2+1-2t+1t8dt [(a-b)2=a2+b2-2ab]

I=t2-2t+2t8dt

I=t2t8-2tt8+2t8dt

I=t2-8-2t1-8+2t-8dt [aman=am-n]

I=t-6-2t-7+2t-8dt

I=t-6dt-2t-7dt+2t-8dt

I=t-6+1-6+1-2t-7+1-7+1+2t-8+1-8+1+C [xndx=xn+1n+1+C]

I=t-5-5-2t-6-6+2t-7-7+C

I=-15t5+26t6-27t7+C

I=-15t5+13t6-27t7+C

Substituting t=1+tanx, we get,

I=-151+tanx5+131+tanx6-271+tanx7+C

Hence, the value of I is -151+tanx5+131+tanx6-271+tanx7+C.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Extrema
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon