4.0 Evaluation By Substitution

In this method we can substitute some function in place of $x$, and change the upper and lower limit according to the new variable.

This method is used to make integral solvable.

Question 3. $I = \int\limits_{{1 \over 3}}^1 {{{{{\left( {x - {x^3}} \right)}^{{1 \over 3}}}} \over {{x^4}}}dx} $

Solution: $$I = \int\limits_{{1 \over 3}}^1 {{{{{\left( {{1 \over {{x^2}}} - 1} \right)}^{{1 \over 3}}}x} \over {{x^4}}}dx} $$ $$I = \int\limits_{{1 \over 3}}^1 {{{{{\left( {{1 \over {{x^2}}} - 1} \right)}^{{1 \over 3}}}} \over {{x^3}}}dx} $$ Let ${1 \over {{x^2}}} - 1 = t$ $$ - {2 \over {{x^3}}}dx = dt$$ $$x = {1 \over 3} \Rightarrow t = 8$$ and $$x = 1 \Rightarrow t = 0$$ $$I = - {1 \over 2}\int\limits_8^0 {{t^{{1 \over 3}}}dt} $$ $$I = - {1 \over 2}\left[ {{{{t^{{1 \over 3} + 1}}} \over {{1 \over 3} + 1}}} \right]_8^0$$ $$I = - {1 \over 2} \cdot {3 \over 4}\left[ {{t^{{4 \over 3}}}} \right]_8^0$$ $$I = - {1 \over 2} \cdot {3 \over 4}\left[ {0 - {8^{{4 \over 3}}}} \right]$$ $$I = {{3 \times 16} \over 8} = 6$$

Question 4. $I = \int\limits_0^1 {{e^{\ln {{\tan }^{ - 1}}x}} \cdot {{\sin }^{ - 1}}\left( {\cos x} \right)dx} $

Solution: $$I = \int\limits_0^1 {{{\tan }^{ - 1}}x \cdot {{\sin }^{ - 1}}\left( {\sin \left( {{\pi \over 2} - x} \right)} \right)dx} $$ $$I = \int\limits_0^1 {{{\tan }^{ - 1}}x \cdot \left( {{\pi \over 2} - x} \right)dx} $$ $$I = {\pi \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}xdx} - \int\limits_0^1 {x{{\tan }^{ - 1}}xdx} $$ By Integration By Parts, $$I = {\pi \over 2}\left[ {\left( {x{{\tan }^{ - 1}}x} \right)_0^1 - \int\limits_0^1 {{x \over {1 + {x^2}}}dx} } \right] - \left[ {\left( {{{{x^2}} \over 2}{{\tan }^{ - 1}}x} \right)_0^1 - {1 \over 2}\int\limits_0^1 {{{{x^2}} \over {1 + {x^2}}}dx} } \right]$$ $$I = {\pi \over 2}\left[ {\left( {x{{\tan }^{ - 1}}x} \right)_0^1 - {1 \over 2}\left( {\ln \left( {1 + {x^2}} \right)} \right)_0^1} \right] - \left[ {\left( {{{{x^2}} \over 2}{{\tan }^{ - 1}}x} \right)_0^1 - {1 \over 2}\int\limits_0^1 {{{{x^2}} \over {1 + {x^2}}}dx} } \right]$$ In the fourth integral, let $x = \tan t$ $$dx = {\sec ^2}tdt$$ $$I = {\pi \over 2}\left[ {\left( {x{{\tan }^{ - 1}}x} \right)_0^1 - {1 \over 2}\left( {\ln \left( {1 + {x^2}} \right)} \right)_0^1} \right] - \left[ {\left( {{{{x^2}} \over 2}{{\tan }^{ - 1}}x} \right)_0^1 - {1 \over 2}\int\limits_0^{{\pi \over 4}} {{{{{\tan }^2}t{{\sec }^2}t} \over {1 + {{\tan }^2}t}}dt} } \right]$$ $$I = {\pi \over 2}\left[ {\left( {x{{\tan }^{ - 1}}x} \right)_0^1 - {1 \over 2}\left( {\ln \left( {1 + {x^2}} \right)} \right)_0^1} \right] - \left[ {\left( {{{{x^2}} \over 2}{{\tan }^{ - 1}}x} \right)_0^1 - {1 \over 2}\int\limits_0^{{\pi \over 4}} {{{{{\tan }^2}t{{\sec }^2}t} \over {{{\sec }^2}t}}dt} } \right]$$ $$I = {\pi \over 2}\left[ {\left( {x{{\tan }^{ - 1}}x} \right)_0^1 - {1 \over 2}\left( {\ln \left( {1 + {x^2}} \right)} \right)_0^1} \right] - \left[ {\left( {{{{x^2}} \over 2}{{\tan }^{ - 1}}x} \right)_0^1 - {1 \over 2}\int\limits_0^{{\pi \over 4}} {\left( {{{\sec }^2}t - 1} \right)dt} } \right]$$ $$I = {\pi \over 2}\left[ {\left( {x{{\tan }^{ - 1}}x} \right)_0^1 - {1 \over 2}\left( {\ln \left( {1 + {x^2}} \right)} \right)_0^1} \right] - \left[ {\left( {{{{x^2}} \over 2}{{\tan }^{ - 1}}x} \right)_0^1 - {1 \over 2}\left( {\tan t - t} \right)_0^{{\pi \over 4}}} \right]$$ $$I = {\pi \over 2}\left[ {{\pi \over 4} - 0 - {1 \over 2}\ln 2 + 0} \right] - \left[ {{\pi \over 8} - 0 - {1 \over 2} + {\pi \over 8}} \right]$$ $$I = {{{\pi ^2}} \over 8} - {\pi \over 4}\ln 2 - {\pi \over 4} + {1 \over 2}$$