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Question: Repeated application of integration by parts gives us the reduction formula, if the integrand is dep...

Repeated application of integration by parts gives us the reduction formula, if the integrand is dependent on a natural number n.
If∫cos⁡mxsin⁡nxdx=cos⁡m−1x(m−n)sin⁡n−1x+A∫cos⁡m−2xsin⁡nxdx+CIf\int {\dfrac{{{{\cos }^m}x}}{{{{\sin }^n}x}}dx = \dfrac{{{{\cos }^{m - 1}}x}}{{(m - n){{\sin }^{n - 1}}x}}} + A\int {\dfrac{{{{\cos }^{m - 2}}x}}{{{{\sin }^n}x}}} dx + C, then A is equal to:
A) mm+n\dfrac{m}{{m + n}}
B) m−1m+n\dfrac{{m - 1}}{{m + n}}
C) mm+n−1\dfrac{m}{{m + n - 1}}
D) m−1m−n\dfrac{{m - 1}}{{m - n}}

Explanation

Solution

According to question given above first of all we have to solve the integration as given in the question,
∫cos⁡mxsin⁡nxdx\int {\dfrac{{{{\cos }^m}x}}{{{{\sin }^n}x}}} dx and to solve the given integration we will let I=∫cos⁡mxsin⁡nxdxI = \int {\dfrac{{{{\cos }^m}x}}{{{{\sin }^n}x}}} dx. Now to find the integration of the integrand first of all we have let sin⁡nx=t{\sin ^n}x = tand we would have to find the differentiation of both of the sides with respect to x. We can also understand it with the help of the formula given below:
d(sin⁡x)dx=cos⁡x...............................(a)\dfrac{{d(\sin x)}}{{dx}} = \cos x...............................(a)
Now, we will substitute the value of dx in the obtained equation and find the integration with respect to t. But, as obtained our integrands are in the products of two functions/terms. So, to find the integration of those two functions/terms we will use the by-step method but first we have to choose the first and the second terms to use the formula given below so that we can find the integration of those terms by step method.

Formula used:
∫f(x)g(x)dx=f(x)∫g(x)dx−∫(df(x)dx∫g(x)dx)d(x)........................(b)\int {f(x)g(x)dx = f(x)\int {g(x)dx - \int {\left( {\dfrac{{df(x)}}{{dx}}\int {g(x)dx} } \right)d(x)........................} } } (b)
Where,
f(x)f(x)is the first function/term and,
g(x)g(x)is the second function/term.
But, before using the formula we have to decide which function/term we should choose as first function/term and second function/term and to choose the first function/term we have to use the LIATE rule which can explained as:
L – Logs
I – Inverse
A – Algebraic
T – Trig
E – Exponential
∫andx=nan−1.................................(c)\int {{a^n}dx = n{a^{n - 1}}} .................................(c)
ddx(a)n=−n(a)n−2.............................(d)\dfrac{d}{{dx{{(a)}^n}}} = - n{(a)^{n - 2}}.............................(d)
On comparing the obtained integration with the given integration in the question we can find the value of A.

Complete step by step answer:
Step 1: To find the value of A first of all we will let,
sin⁡nx=t{\sin ^n}x = t………………………(2)
Step 2: Now, we will differentiate the both sides of the equation (2) with respect to x, with the help of the formula (a) as mentioned in the solution hint.
d(sin⁡x)dx=dtdx\dfrac{{d(\sin x)}}{{dx}} = \dfrac{{dt}}{{dx}}
Hence,
cos⁡xdx=dt ⇒dx=dtcos⁡x.......................(3)  \cos xdx = dt \\\ \Rightarrow dx = \dfrac{{dt}}{{\cos x}}.......................(3) \\\
Step 3: Now, we will take cos⁡x\cos xas a common term from the given integration I=∫cos⁡mxsin⁡nxdxI = \int {\dfrac{{{{\cos }^m}x}}{{{{\sin }^n}x}}dx} which is given in the question.
As we know, cos⁡x×cos⁡m−1x=cos⁡mx\cos x \times {\cos ^{m - 1}}x = {\cos ^m}x
Hence,
⇒I=∫cos⁡xcos⁡m−1xsin⁡nxdx\Rightarrow I = \int {\dfrac{{\cos x{{\cos }^{m - 1}}x}}{{{{\sin }^n}x}}dx} ……………………..(4)
Step 4: On substituting the value of dxdx from equation (3) in equation (4)
⇒I=∫cos⁡xcos⁡m−1xdtsin⁡nxcos⁡x\Rightarrow I = \int {\dfrac{{\cos x{{\cos }^{m - 1}}xdt}}{{{{\sin }^n}x\cos x}}}
As, the equation obtained above we will eliminate the cos⁡x\cos xso the after elimination the equation will be:
⇒I=∫cos⁡m−1xdtsin⁡nx\Rightarrow I = \int {\dfrac{{{{\cos }^{m - 1}}xdt}}{{{{\sin }^n}x}}}
As we know that,
cos⁡2x=1−sin⁡2x{\cos ^2}x = 1 - {\sin ^2}x
Hence,
⇒I=∫(1−sin⁡2x)m−12dtsin⁡nx\Rightarrow I = \int {\dfrac{{{{(1 - {{\sin }^2}x)}^{\dfrac{{m - 1}}{2}}}dt}}{{{{\sin }^n}x}}}
Step 5: Now from equation (2), we will substitute the value of sin⁡x\sin xas we let in the step 1 of the solution.
⇒I=∫(1−t2)m−12dttn\Rightarrow I = \int {\dfrac{{{{(1 - {t^2})}^{\dfrac{{m - 1}}{2}}}dt}}{{{t^n}}}}
On, rearranging the numerator of the integrand which is (1−t2)m−12{(1 - {t^2})^{\dfrac{{m - 1}}{2}}}
(1−t2)m−12=tm−1(1t2−1)m−12{(1 - {t^2})^{\dfrac{{m - 1}}{2}}} = {t^{m - 1}}{\left( {\dfrac{1}{{{t^2}}} - 1} \right)^{\dfrac{{m - 1}}{2}}}
Therefore,
⇒I=∫tm−1(1t2−1)m−12tndt\Rightarrow I = \int {\dfrac{{{t^{m - 1}}{{\left( {\dfrac{1}{{{t^2}}} - 1} \right)}^{\dfrac{{m - 1}}{2}}}}}{{{t^n}}}} dt
Now, on dividing tm−1(1t2−1)m−12{t^{m - 1}}{\left( {\dfrac{1}{{{t^2}}} - 1} \right)^{\dfrac{{m - 1}}{2}}}by tn{t^n}
⇒I=∫tm−1tn(1t2−1)m−121dt\Rightarrow I = \int {\dfrac{{\dfrac{{{t^{m - 1}}}}{{{t^n}}}{{\left( {\dfrac{1}{{{t^2}}} - 1} \right)}^{\dfrac{{m - 1}}{2}}}}}{1}} dt
I=tm−n−1(1t2−1)m−12dtI = {t^{m - n - 1}}{\left( {\dfrac{1}{{{t^2}}} - 1} \right)^{\dfrac{{m - 1}}{2}}}dt
Step 6: Now, we have to solve the integration of the obtained integrand by using the by-part integration method as mentioned in the solution hint.
So, according to the by-part rule we will take (1t2−1)m−12{\left( {\dfrac{1}{{{t^2}}} - 1} \right)^{\dfrac{{m - 1}}{2}}}as the first term and tm−n−1{t^{m - n - 1}}as the second function.
Step 7: Now, to solve the obtained terms of the integration we use the formula (b) as mentioned in the solution hint.
⇒I=(1t2−1)m−12∫tm−n−1dt−∫[ddt(1t2−1)m−12∫tm−n−1dt]dt   \Rightarrow I = {\left( {\dfrac{1}{{{t^2}}} - 1} \right)^{\dfrac{{m - 1}}{2}}}\int {{t^{m - n - 1}}dt - \int {\left[ {\dfrac{d}{{dt}}{{\left( {\dfrac{1}{{{t^2}}} - 1} \right)}^{\dfrac{{m - 1}}{2}}}\int {{t^{m - n - 1}}dt} } \right]dt} } \\\ \\\
Now we can solve the integration with the help of the formula (c) and (d).
Hence,
⇒I=(1t2−1)m−12tm−nm−n−∫(m−12(1t2−1)m−32(−2t3)tm−nm−n)dt\Rightarrow I = {\left( {\dfrac{1}{{{t^2}}} - 1} \right)^{\dfrac{{m - 1}}{2}}}\dfrac{{{t^{m - n}}}}{{m - n}} - \int {\left( {\dfrac{{m - 1}}{2}{{\left( {\dfrac{1}{{{t^2}}} - 1} \right)}^{\dfrac{{m - 3}}{2}}}\left( {\dfrac{{ - 2}}{{{t^3}}}} \right)\dfrac{{{t^{m - n}}}}{{m - n}}} \right)dt} ……………………………………(5)
Step 8: On substituting the value of t from the equation 1 which is sin⁡x\sin xin equation (5)
⇒I=(1(sin⁡x)2−1)m−12(sin⁡x)m−nm−n−∫(m−12(1(sin⁡x)2−1)m−32(−2(sin⁡x)3)(sin⁡x)m−nm−ncos⁡x)dx\Rightarrow I = {\left( {\dfrac{1}{{{{(\sin x)}^2}}} - 1} \right)^{\dfrac{{m - 1}}{2}}}\dfrac{{{{(\sin x)}^{m - n}}}}{{m - n}} - \int {\left( {\dfrac{{m - 1}}{2}{{\left( {\dfrac{1}{{{{(\sin x)}^2}}} - 1} \right)}^{\dfrac{{m - 3}}{2}}}\left( {\dfrac{{ - 2}}{{{{(\sin x)}^3}}}} \right)\dfrac{{{{(\sin x)}^{m - n}}}}{{m - n}}\cos x} \right)} dx
On solving the obtained equation,

⇒I=(1−sin⁡2xsin⁡2x)m−12(sin⁡x)m−nm−n−∫[m−12(1−sin⁡2xsin⁡2x)m−32(−2(sin⁡x)3(sin⁡x)m−nm−ncos⁡x)]dx ⇒I=(cos⁡2xsin⁡2x)m−12(sin⁡x)m−nm−n−∫[m−12(cos⁡2xsin⁡2x)m−32(−2(sin⁡x)3(sin⁡x)m−nm−ncos⁡x)]dx  \Rightarrow I = {\left( {\dfrac{{1 - {{\sin }^2}x}}{{{{\sin }^2}x}}} \right)^{\dfrac{{m - 1}}{2}}}\dfrac{{{{(\sin x)}^{m - n}}}}{{m - n}} - {\int {[\dfrac{{m - 1}}{2}\left( {\dfrac{{1 - {{\sin }^2}x}}{{{{\sin }^2}x}}} \right)} ^{\dfrac{{m - 3}}{2}}}\left( {\dfrac{{ - 2}}{{{{(\sin x)}^3}}}\dfrac{{{{(\sin x)}^{m - n}}}}{{m - n}}\cos x} \right)]dx \\\ \Rightarrow I = {\left( {\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}} \right)^{\dfrac{{m - 1}}{2}}}\dfrac{{{{(\sin x)}^{m - n}}}}{{m - n}} - {\int {[\dfrac{{m - 1}}{2}\left( {\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}} \right)} ^{\dfrac{{m - 3}}{2}}}\left( {\dfrac{{ - 2}}{{{{(\sin x)}^3}}}\dfrac{{{{(\sin x)}^{m - n}}}}{{m - n}}\cos x} \right)]dx \\\

Step 9: After the simplification of the obtained equation,
⇒I=cos⁡m−1x(m−n)sin⁡n−1x+m−1m−n∫cos⁡m−2xsin⁡nxdx\Rightarrow I = \dfrac{{{{\cos }^{m - 1}}x}}{{(m - n){{\sin }^{n - 1}}x}} + \dfrac{{m - 1}}{{m - n}}\int {\dfrac{{{{\cos }^{m - 2}}x}}{{{{\sin }^n}x}}dx} ………………………….(6)
Step 10: Now, to find the value of A we would have to compare the (6) with the equation as given in the question: ∫cos⁡mxsin⁡nxdx=cos⁡m−1x(m−n)sin⁡n−1x+A∫cos⁡m−2xsin⁡nxdx+C\int {\dfrac{{{{\cos }^m}x}}{{{{\sin }^n}x}}dx = \dfrac{{{{\cos }^{m - 1}}x}}{{(m - n){{\sin }^{n - 1}}x}}} + A\int {\dfrac{{{{\cos }^{m - 2}}x}}{{{{\sin }^n}x}}} dx + C
Where I=∫cos⁡mxsin⁡nxdxI = \int {\dfrac{{{{\cos }^m}x}}{{{{\sin }^n}x}}dx}
⇒cos⁡m−1x(m−n)sin⁡n−1x+A∫cos⁡m−2xsin⁡nxdx=cos⁡m−1x(m−n)sin⁡n−1x+m−1m−n∫cos⁡m−2xsin⁡nxdx\Rightarrow \dfrac{{{{\cos }^{m - 1}}x}}{{(m - n){{\sin }^{n - 1}}x}} + A\int {\dfrac{{{{\cos }^{m - 2}}x}}{{{{\sin }^n}x}}} dx = \dfrac{{{{\cos }^{m - 1}}x}}{{(m - n){{\sin }^{n - 1}}x}} + \dfrac{{m - 1}}{{m - n}}\int {\dfrac{{{{\cos }^{m - 2}}x}}{{{{\sin }^n}x}}} dx
Therefore,
⇒A=m−1m−n\Rightarrow A = \dfrac{{m - 1}}{{m - n}}

Hence, by comparing the equation (6) with the equation given in the question we have obtained the value of A which is A=m−1m−nA = \dfrac{{m - 1}}{{m - n}}therefore, option D is correct.

Note:
- If the given integration is a product of two functions/terms then to solve this type of integration we will have to use the Integration by part rule. Where we have to choose the first and second function/term using the LIATE rule where, L belongs to Logs, I belongs to Inverse, A belongs to Algebraic, T belongs to Trig, and E belongs to Exponential.
- If the equation is getting complicated try two solutions in parts.
- The constant expression, allows us to ignore the constant coefficient in the expression while we integrate the rest of the expression.