To determine
To find:
The derivative of the function gx=∫tanxx212+t4 dt
Answer
g'x=-sec2x2+tan4x+2x2+x8
Explanation
1) Concept:
i) Properties of definite integral:
∫ba fxdx=-∫ab fxdx
2) Fundamental theorem of Calculus, Part 1:
If f is continuous on a, b, then the function g defined by
gx=∫0x ft dt a≤x≤b
gx is continuous on a,b, and g'x=f(x)
3) Given:
gx=∫tanxx212+t4 dt
4) Calculation:
Rewrite the given integral as (by using hint given)
∫tanxx212+t4 dt=∫tanx012+t4 dt+∫0x212+t4 dt
Now by using concept i) (Property of definite integral),
=-∫0tanx12+t4 dt+∫0x212+t4 dt
Differentiating with respect to x
g'x=ddx-∫0tanx12+t4 dt+ddx∫0x212+t4 dt
By using Fundamental theorem of Calculus, Part 1,
g'x=-12+t4ddtt- 12+t4ddt0+12+t4ddtt- 12+t4ddt0
Substituting the values of limit (t=tanxand 0 in first term t=x2 and 0 in second term),
g'x=-12+tanx4ddx(tanx)- 12+tanx4ddx0+12+x24ddx(x2)- 12+x24ddx0
=-12+tan4xddxtanx+0+12+x8ddxx2- 0
=-12+(tanx)4sec2x+12+x82x
=-sec2x2+tan4x+2x2+x8
Therefore,
g'x=-sec2x2+tan4x+2x2+x8
Conclusion:
Therefore, the derivative the given integral is
-sec2x2+tan4x+2x2+x8