Class 12 RD Sharma Solutions- Chapter 20 Definite Integrals – Exercise 20.4 Part A

Evaluate each of the following integrals (1-16):

Question 1. $\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx$

Solution:

We know that $\int\limits_0^{2π }f(x)dx=\int\limits_0^{2π }f(2π -x)dx$
so,
$\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx =\int\limits_0^{2π}\frac{e^{sin(2π-x)}}{e^{sin(2π-x)}+e^{-sin(2π-x)}}dx$
we know, $sin(2π -x)=-sinx$
$\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx =\int\limits_0^{2π}\frac{e^{-sinx}}{e^{-sinx}+e^{sinx}}dx$
if
I = $\int\limits_0^{2π}\frac{e^{-sinx}}{e^{-sinx}+e^{sinx}}dx$
then
I = $\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx$
2I = $\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx +\int\limits_0^{2π}\frac{e^{-sinx}}{e^{-sinx}+e^{sinx}}dx$
$2I=\int\limits_0^{2π}(\frac{e^{sinx}}{e^{sinx}+e^{-sinx}} +\frac{e^{-sinx}}{e^{-sinx}+e^{sinx}})dx$
$2I=\int\limits_0^{2π}(\frac{e^{sinx}+e^{-sinx}}{e^{sinx}+e^{-sinx}} )dx$
$2I=\int\limits_0^{2π}dx$
$2I=2π$
l=π

Question 2. $\int\limits_0^{2π} log (secx+tanx)dx$

Solution:

We know that $\int\limits_0^{2π }f(x)dx=\int\limits_0^{2π }f(2π -x)dx$
So,
$\int\limits_0^{2π} log (secx+tanx)dx= \int\limits_0^{2π} log (sec(2π-x)+tan(2π-x)dx$
$\int\limits_0^{2π} log (secx+tanx)dx= \int\limits_0^{2π} log (secx-tanx)dx$
if
I = $\int\limits_0^{2π} log (secx-tanx)dx$
then
I = $\int\limits_0^{2π} log (secx+tanx)dx$
2I = $\int\limits_0^{2π} log (secx+tanx)dx+ \int\limits_0^{2π} log (secx-tanx)dx$
2I= $\int\limits_0^{2π} log (sec^2x-tan^2x)dx$
2I = $\int\limits_0^{2π} log (1)dx$
2I=0
I=0

Question 3. $\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx$

Solution:

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$
So,
$\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx }+\sqrt{cotx}}dx = \int\limits_{π/6}^{π/3}\frac{\sqrt{tan(π/2-x)}}{\sqrt{tan(π/2-x)}+\sqrt{cot(π/2-x)}}dx$
$\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx }+\sqrt{cotx}}dx = \int\limits_{π/6}^{π/3}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx$
if
I= $\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx$
then
I= $\int\limits_{π/6}^{π/3}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx$
So
2I= $\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx +\int\limits_{π/6}^{π/3}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx$
2I = $\int\limits_{π/6}^{π/3}(\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}} +\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}})dx$
2I= $\int\limits_{π/6}^{π/3}dx$
2I=π/6
I=π/12

Question 4. $\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx$

Solution:

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$
So,
$\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx = \int\limits_{π/6}^{π/3} \frac{\sqrt{sin(π/2-x)}}{\sqrt{sin(π/2-x)}+\sqrt{cos(π/2-x)}} dx$
$\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx = \int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}} dx$
if
I = $\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx$
then,
I = $\int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}} dx$
2I = $\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx +\int\limits_{π/6}^{π/3}\frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}} dx$
2I = $\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}+\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}} dx$
2I = $\int\limits_{π/6}^{π/3}dx$
2I=π/6
I=π/12

Question 5. $\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx$

Solution:

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$
so,
$\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx =\int\limits_{-π/4}^{π/4} \frac{tan^2{-x}}{1+e^{-x}}dx$
$\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^{-x}}dx$
if
$I=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx$
then,
$I=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^{-x}}dx$
$2I=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx +\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^{-x}}dx$
$2I=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx +\int\limits_{-π/4}^{π/4} \frac{e^xtan^2x}{1+e^x}dx$
$2I=\int\limits_{-π/4}^{π/4} \frac{tan^2x+e^xtan^2x}{1+e^x}dx$
$2I=\int\limits_{-π/4}^{π/4} \frac{(e^x+1)tan^2x}{1+e^x}dx$
$2I=\int\limits_{-π/4}^{π/4} tan^2xdx$
$I=\frac{\int\limits_{-π/4}^{π/4} tan^2x}{2}dx$
we know if
f(x) is even $\int\limits_{-a}^af(x)dx=2\int\limits_{0}^af(x)dx$
f(x) is odd $\int\limits_{-a}^af(x)dx=0$
Here, f(x) = tan²x which is even
hence,
$I=\int\limits_{0}^{π/4}tan^2xdx$
$I=\int\limits_{0}^{π/4}(sec^2x-1)dx$
$I=\left.({tanx-x})\right|_0^{π/4}$
I = $I=π/4$

Question 6. $\int\limits_{-a}^a \frac{1}{1+a^x}dx,a>0$

Solution:

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$
So,
$\int\limits_{-a}^a \frac{1}{1+a^x}dx = \int\limits_{-a}^a \frac{1}{1+a^{-x}}dx$
if, $I = \int\limits_{-a}^a \frac{1}{1+a^x}dx$ then $I=\int\limits_{-a}^a \frac{1}{1+a^{-x}}dx$
So,
$2I=\int\limits_{-a}^a \frac{1}{1+a^x}dx+ \int\limits_{-a}^a \frac{1}{1+a^{-x}}dx$
$2I=\int\limits_{-a}^a \frac{1}{1+a^x}+ \frac{1}{1+a^{-x}}dx$
$2I= \int\limits_{-a}^a \frac{1+a^x}{1+a^x}dx$
$2I= \int\limits_{-a}^a \frac{1}{1+a^x}+ \frac{a^x}{1+a^x}dx$
$2I= \int\limits_{-a}^a dx$
$2I=2a$
$I=a$

Question 7. $\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx$

Solution:

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$
Hence,
$\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx =\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{-tanx}} dx$
if, $I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx$
then $I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{-tanx}} dx$
so,
$2I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx+\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{-tanx}} dx$
$2I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} + \frac{1}{1+e^{-tanx}} dx$
$2I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} + \frac{e^{tanx}}{1+e^{tanx}} dx$
$2I=\int\limits_{-π/3}^{π/3} dx$
$2I=2π/3$
$I=π/3$

Question 8. $\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx$

Solution:

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$
hence, $\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx =\int\limits_{-π/2}^{π/2} \frac{cos^2(-x)}{1+e^{-x}}dx$
$\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx = \int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^{-x}}dx$
if $I=\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx$
Then, $I=\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^{-x}}dx$
So, $2I=\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx +\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^{-x}}dx$
$2I= \int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}+ \frac{e^xcos^2x}{1+e^x}dx$
$2I=\int\limits_{-π/2}^{π/2} \frac{cos^2x+e^xcos^2x}{1+e^x}dx$
$2I=\int\limits_{-π/2}^{π/2} \frac{(1+e^x)cos^2x}{1+e^x}dx$
$2I=\int\limits_{-π/2}^{π/2} cos^2x dx$
$2I=\int\limits_{-π/2}^{π/2} \frac{1+cos2x}{2}dx$
$I= \left.\frac{1}{4}(x+\frac{sin2x}2)\right|_{-π/2}^{π/2}$
$I=\frac{1}{4}[\frac{π}2-(-\frac{π}2)]$
$I=\frac{π}4$

Question 9. $\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5+1}{cos^2x} dx$

Solution:

$\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5+1}{cos^2x} dx$
$\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5}{cos^2x}+\frac{1}{cos^2x} dx$
$\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5}{cos^2x}+\int\limits_{-π/4}^{π/4}sec^2x dx$
if f(x) is even
$\int\limits_{-a}^af(x)dx=2\int\limits_{0}^af(x)dx$
if f(x) is odd
$\int\limits_{-a}^af(x)dx=0$
here, $\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5}{cos^2x}dx$ is odd and $\int\limits_{-π/4}^{π/4}sec^2x dx$
is even
Hence,
$0+2\int\limits_0^{π/4}sec^2x dx$
$\left.2tanx\right|_0^{\frac{π}{4}}$
2

Question 10. $\int\limits_{a}^{b} \frac{x^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx,n\in N,n\geq 2$

Solution:

if
$I=\int\limits_{a}^{b} \frac{x^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx$
then,
$I=\int\limits_{a}^{b} \frac{{(a+b-x)}^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx$
$2I=\int\limits_{a}^{b} \frac{x^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx +\int\limits_{a}^{b} \frac{(a+b-x)^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx$
$2I=\int\limits_{a}^{b} \frac{x^{1/n}+(a+b-x)^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx$
$2I=\int\limits_{a}^{b} dx$
$I=\frac{b-a}{2}$

Question 11. $\int\limits_0^{π/2} (2logcosx-logsinx2x)dx$

Solution:

let, $I=\int\limits_0^{π/2} (2logcosx-logsinx2x)dx$
$I=\int\limits_0^{π/2} (log\frac{cos^2x}{sinx2x})dx$
$I=\int\limits_0^{π/2} (log\frac{cos^2x}{2sinxcosx})dx$
$I=\int\limits_0^{π/2} (log\frac{cosx}{2sinx})dx$
$I=\int\limits_0^{π/2} logcosxdx-\int\limits_0^{π/2} logsinxdx-\int\limits_0^{π/2}log2dx$
we know that, $\int\limits_0^{π/2}logcosxdx=\int\limits_0^{π/2}logsinxdx$
hence,
$I=-\int\limits_0^{π/2}log2dx =\frac{-π}{2}log2$

Question 12. $\int\limits_0^a \frac{\sqrt x}{\sqrt x+\sqrt {a-x}} dx$

Solution:

Let, $I=\int\limits_0^a \frac{\sqrt x}{\sqrt x+\sqrt {a-x}} dx$
we know that , $\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx$
so, $I=\int\limits_0^a \frac{\sqrt {a-x}}{\sqrt{a-x}+\sqrt {x}} dx$
then,
$2I= \int\limits_0^a \frac{\sqrt x}{\sqrt x+\sqrt {a-x}} dx + \int\limits_0^a \frac{\sqrt{ a-x}}{\sqrt {a-x}+\sqrt x} dx$
$2I= \int\limits_0^a \frac{\sqrt x+\sqrt{a-x}}{\sqrt x+\sqrt {a-x}} dx$
$2I= \int\limits_0^a dx$
$2I=a$
$I=\frac{a}{2}$

Question 13. $\int\limits_0^5 \frac{\sqrt[4]{x+4} }{\sqrt[4]{x+4} +\sqrt[4]{9-x} }dx$

Solution:

We know that, $\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx$
So,
$I=\int\limits_0^5 \frac {\sqrt[4]{(5-x)+4} }{\sqrt[4]{(5-x)+4} +\sqrt[4]{9-(5-x)} }dx$
$I=\int\limits_0^5 \frac{\sqrt[4]{9-x} }{\sqrt[4]{9-x} +\sqrt[4]{4+x} }dx$
then,
$2I=\int\limits_0^5 \frac{\sqrt[4]{x+4} }{\sqrt[4]{x+4} +\sqrt[4]{9-x} }dx +\int\limits_0^5 \frac{\sqrt[4]{9-x} }{\sqrt[4]{9-x} +\sqrt[4]{4+x} }dx$
$2I= \int\limits_0^5 \frac{\sqrt[4]{x+4}\sqrt[4]{9-x} }{\sqrt[4]{x+4} +\sqrt[4]{9-x} }dx$
$2I= \int\limits_0^5 dx$
I $I=\frac{5}{2}$

Question 14. $\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{7-x}}dx$

Solution:

We know that, $\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx$
so,
$I=\int\limits_0^7 \frac{\sqrt[3]{7-x}}{\sqrt[3]{7-x}+\sqrt[3]{x}}dx$
then,
$2I=\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{7-x}}dx+ \int\limits_0^7 \frac{\sqrt[3]{7-x}}{\sqrt[3]{7-x}+\sqrt[3]{x}}dx$
$2I=\int\limits_0^7 dx$
$I=\frac{7}{2}$

Question 15. $\int\limits_{π/6}^{π/3} \frac{1}{1+\sqrt {tanx}}dx$

Solution:

We know that, $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$
Let, $I=\int\limits_{π/6}^{π/3} \frac{1}{1+\sqrt {tanx}}dx$
$I=\int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt {sinx}}dx$
hence,
$I=\int\limits_{π/6}^{π/3} \frac{\sqrt{cos(\frac{π}{2}- x)}}{\sqrt{cos(\frac{π}{2}-x)} +\sqrt {sin(\frac{π}{2}-x)}}dx$
$I=\int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt {sinx}}dx$
$2I=\int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt {sinx}}dx + \int\limits_{π/6}^{π/3} \frac{\sqrt{sinx}}{\sqrt{cosx}+\sqrt {sinx}}dx$
$2I= \int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}+\sqrt{sinx}}{\sqrt{cosx}+\sqrt {sinx}}dx$
$2I= \int\limits_{π/6}^{π/3} dx$
$I=\frac{π}{12}$

Question 16. If f(a+b-x)=f(x), then prove that $\int\limits_a^b xf(x)dx=\frac{a+b}{2} \int\limits_a^bf(x)dx$

Solution:

$I=\int\limits_a^bf(x)dx$
$I=\int\limits_a^b(a+b-x)f(a+b-x)dx$
$I=\int\limits_a^b(a+b-x)f(x)dx........... [\because f(a+b-x)=f(x)]$
$I=\int\limits_a^b(a+b)f(x)dx-\int\limits_a^bf(x)dx$
$I=(a+b)\int\limits_a^bf(x)dx-I$
$2I=(a+b)\int\limits_a^bf(x)dx$
$I=\frac{(a+b)}{2}\int\limits_a^bf(x)dx$
$\therefore \int\limits_a^bxf(x)dx=\frac{(a+b)}{2}\int\limits_a^bf(x)dx$

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Last Updated : 22 Dec, 2022

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Class 12 RD Sharma Solutions- Chapter 20 Definite Integrals – Exercise 20.4 Part A

Evaluate each of the following integrals (1-16):

Question 1. $\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx$

Question 2. $\int\limits_0^{2π} log (secx+tanx)dx$

Question 3. $\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx$

Question 4. $\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx$

Question 5. $\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx$

Question 6. $\int\limits_{-a}^a \frac{1}{1+a^x}dx,a>0$

Question 7. $\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx$

Question 8. $\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx$

Question 9. $\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5+1}{cos^2x} dx$

Question 10. $\int\limits_{a}^{b} \frac{x^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx,n\in N,n\geq 2$

Question 11. $\int\limits_0^{π/2} (2logcosx-logsinx2x)dx$

Question 12. $\int\limits_0^a \frac{\sqrt x}{\sqrt x+\sqrt {a-x}} dx$

Question 13. $\int\limits_0^5 \frac{\sqrt[4]{x+4} }{\sqrt[4]{x+4} +\sqrt[4]{9-x} }dx$

Question 14. $\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{7-x}}dx$

Question 15. $\int\limits_{π/6}^{π/3} \frac{1}{1+\sqrt {tanx}}dx$

Question 16. If f(a+b-x)=f(x), then prove that $\int\limits_a^b xf(x)dx=\frac{a+b}{2} \int\limits_a^bf(x)dx$

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

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Class 12 RD Sharma Solutions- Chapter 20 Definite Integrals – Exercise 20.4 Part A

Evaluate each of the following integrals (1-16):

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

Question 13.

Question 14.

Question 15.

Question 16. If f(a+b-x)=f(x), then prove that

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Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Question 1. $\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx$

Question 2. $\int\limits_0^{2π} log (secx+tanx)dx$

Question 3. $\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx$

Question 4. $\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx$

Question 5. $\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx$

Question 6. $\int\limits_{-a}^a \frac{1}{1+a^x}dx,a>0$

Question 7. $\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx$

Question 8. $\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx$

Question 9. $\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5+1}{cos^2x} dx$

Question 10. $\int\limits_{a}^{b} \frac{x^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx,n\in N,n\geq 2$

Question 11. $\int\limits_0^{π/2} (2logcosx-logsinx2x)dx$

Question 12. $\int\limits_0^a \frac{\sqrt x}{\sqrt x+\sqrt {a-x}} dx$

Question 13. $\int\limits_0^5 \frac{\sqrt[4]{x+4} }{\sqrt[4]{x+4} +\sqrt[4]{9-x} }dx$

Question 14. $\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{7-x}}dx$

Question 15. $\int\limits_{π/6}^{π/3} \frac{1}{1+\sqrt {tanx}}dx$

Question 16. If f(a+b-x)=f(x), then prove that $\int\limits_a^b xf(x)dx=\frac{a+b}{2} \int\limits_a^bf(x)dx$