MATH101 April 2016
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q2 (a) • Q10 (a) • Q10 (b) • Q11 (a) • Q11 (b) • Q2 (b) • Q2 (c) • Q2 (d) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q7 • Q8 (a) • Q8 (b) • Q9 (a) • Q9 (b) •
Question 01 (c)

Which integral equals $\displaystyle \int _{0}^{\pi /2}f(\sin x)\,dx$?
${\begin{aligned}{{\textbf {J}}{:}}&\,\int _{0}^{1}f(u)\,du&{{\textbf {M}}{:}}&\,\int _{0}^{\arcsin(\pi /2)}f(u)\,du&{{\textbf {Q}}{:}}&\,\int _{0}^{\pi /2}f(u)\,du\\{{\textbf {K}}{:}}&\,\int _{0}^{1}{\frac {f(u)}{\sqrt {1u^{2}}}}\,du&{{\textbf {N}}{:}}&\,\int _{0}^{\arcsin(\pi /2)}{\frac {f(u)}{\sqrt {1u^{2}}}}\,du&{{\textbf {R}}{:}}&\,\int _{0}^{\pi /2}{\frac {f(u)}{\sqrt {1u^{2}}}}\,du\\{{\textbf {L}}{:}}&\,\int _{0}^{1}f(u)\cos u\,du&{{\textbf {P}}{:}}&\,\int _{0}^{\arcsin(\pi /2)}f(u)\cos u\,du&{{\textbf {S}}{:}}&\,\int _{0}^{\pi /2}f(u)\cos u\,du\end{aligned}}$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Putting $u=sinx$, try to apply substitution method.

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Solution

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The idea is using substitution with ${\color {OrangeRed}u=sinx}$.
Since $u=sinx$, we have
$du=cosxdx={\sqrt {1\sin ^{2}x}}dx={\sqrt {1u^{2}}}dx$,
so that ${\color {Maroon}dx={\frac {1}{\sqrt {1u^{2}}}}du}$.
On the other hand, $\sin x_{x=0}=0$ and $\sin x_{x={\frac {\pi }{2}}}=1$, which implies the new integration interval is $[0,1]$.
Therefore, the given integral can be rewritten as
$\int _{0}^{\frac {\pi }{2}}f({\color {OrangeRed}\sin x}){\color {Maroon}dx}=\int _{0}^{1}f({\color {OrangeRed}u}){\color {Maroon}{\frac {1}{\sqrt {1u^{2}}}}du}=\int _{0}^{1}{\frac {f(u)}{\sqrt {1u^{2}}}}du$.
The answer is K.
