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Question: Evaluate the given integral. \(\int {{{\sec }^{ - 1}}\sqrt {\text{x}} } {\text{dx}}\)...

Evaluate the given integral.
∫sec⁡−1xdx\int {{{\sec }^{ - 1}}\sqrt {\text{x}} } {\text{dx}}

Explanation

Solution

Hint: To solve this question we will let x = t\sqrt {\text{x}} {\text{ = t}} and then we will apply by – parts to find the value of given integration.

Complete step-by-step solution -

Now, we have I = ∫sec⁡−1xdx\int {{{\sec }^{ - 1}}\sqrt {\text{x}} } {\text{dx}}
Let x = t\sqrt {\text{x}} {\text{ = t}}
Therefore, we get xdx = 2tdt{\text{xdx = 2tdt}}.
So, our integration becomes I = 2∫tsec−1tdt2\int {{\text{tse}}{{\text{c}}^{ - 1}}{\text{t}}} {\text{dt}}
Now, we will use integration by – parts to solve the above integration. By – parts is used to find integration of a function which is the multiplication of two other functions. The formula of by – parts is
∫f(x).g(x)dx = f(x)∫g(x)dx - ∫f’(x)(∫g(x)dx)dx\int {{\text{f(x)}}{\text{.g(x)dx = f(x)}}\int {{\text{g(x)dx}}} {\text{ - }}\int {{\text{f'}}({\text{x)}}\left( {\int {{\text{g(x)dx}}} } \right){\text{dx}}} }
Where f(x) is the first function and g(x) is the second function. We will decide the first and second function according to the ILATE rule where ILATE is inverse, logarithmic, algebraic, trigonometric and exponential.
Therefore, to find I = 2∫tsec−1tdt2\int {{\text{tse}}{{\text{c}}^{ - 1}}{\text{t}}} {\text{dt}}, the first function is sec⁡−1t{\sec ^{ - 1}}{\text{t}} and second function is t. So, we will apply by – parts,
I = 2[sec⁡−1t∫tdt - ∫d(sec−1t)dt(∫tdt)dt]2\left[ {{{\sec }^{ - 1}}{\text{t}}\int {{\text{tdt}}} {\text{ - }}\int {\dfrac{{{\text{d(se}}{{\text{c}}^{ - 1}}{\text{t)}}}}{{{\text{dt}}}}} (\int {{\text{tdt}}} ){\text{dt}}} \right]
Now, ∫xndx = xn + 1n + 1\int {{{\text{x}}^{\text{n}}}{\text{dx = }}\dfrac{{{{\text{x}}^{{\text{n + 1}}}}}}{{{\text{n + 1}}}}} and d(sec - 1t)dt = 1tt2 - 1\dfrac{{{\text{d(se}}{{\text{c}}^{{\text{ - 1}}}}{\text{t)}}}}{{{\text{dt}}}}{\text{ = }}\dfrac{1}{{{\text{t}}\sqrt {{{\text{t}}^2}{\text{ - 1}}} }}
So, I = 2[t22sec⁡−1t - ∫1tt2 - 1(t22)dt]2\left[ {\dfrac{{{{\text{t}}^2}}}{2}{{\sec }^{ - 1}}{\text{t - }}\int {\dfrac{1}{{{\text{t}}\sqrt {{{\text{t}}^2}{\text{ - 1}}} }}\left( {\dfrac{{{{\text{t}}^2}}}{2}} \right)} {\text{dt}}} \right]
I = t2sec⁡−1t - ∫tt2 - 1dt{{\text{t}}^2}{\sec ^{ - 1}}{\text{t - }}\int {\dfrac{{\text{t}}}{{\sqrt {{{\text{t}}^2}{\text{ - 1}}} }}} {\text{dt}}
Let t2 - 1 = z{{\text{t}}^2}{\text{ - 1 = z}}
So, 2tdt = dz{\text{2tdt = dz}}
I = t2sec⁡−1t - ∫dz2z{{\text{t}}^2}{\sec ^{ - 1}}{\text{t - }}\int {\dfrac{{{\text{dz}}}}{{2\sqrt {\text{z}} }}}
I = t2sec⁡−1t - z + C{{\text{t}}^2}{\sec ^{ - 1}}{\text{t - }}\sqrt {\text{z}} {\text{ + C}}
So, I = t2sec⁡−1t - t2 - 1 + C{{\text{t}}^2}{\sec ^{ - 1}}{\text{t - }}\sqrt {{{\text{t}}^2}{\text{ - 1}}} {\text{ + C}}
Now, putting value of t, we get
I = xsec⁡−1x - x - 1 + C{\text{x}}{\sec ^{ - 1}}\sqrt {\text{x}} {\text{ - }}\sqrt {{\text{x - 1}}} {\text{ + C}}

Note: When we come up with such types of questions, we have to use integration by – parts to evaluate the given integral. While using by – parts we have to use the ILATE rule. The ILATE is exactly not a rule but it came after many trial and error experiments. Generally, we take those functions as the first function whose integration is not known to us. In the above question, we don’t know the integration of sec⁡−1t{\sec ^{ - 1}}{\text{t}}, so we take it as the first function and t as the second function.