Reduction Formulae-Previous year questions

Previous Year Questions on Reduction Formulae

Reduction Formulae is a fundamental concept in Mathematics, widely studied in Integral Calculus. Historically, reduction formulae have streamlined complex integrations, enabling efficient problem-solving in mathematical analysis. Their importance spans academic studies and real-world applications, ensuring relevance for students. Review Mathematics questions to deepen your knowledge and enhance your problem-solving abilities.

Vidyasagar University

2023-24 (NEP)

If \(m,n\) being positive greater than \(1\), derive the following reduction formula \begin{align*} I_{m,n}&=\int \sin^{m}{x}\cos^{n}{x}\, dx\\ &=\frac{\sin^{m+1}{x}\cos^{n-1}{x}}{m+1}+\frac{n-1}{m+1}I_{m,n-2}-\frac{n-1}{m+1}I_{m,n} ~~~[2] \end{align*}
If \begin{align*} I_{m,n}=\int^{\frac{\pi}{2}}_{0}\sin^{m}{x}\cos^{n}{x} \, dx \end{align*} \(m,n\) being positive integers greater than \(1\), prove that then show that \begin{align*} I_{m,n}=\frac{n-1}{m+n}I_{m,n-2} \end{align*} Hence find the value of \begin{align*} \int^{1}_{0}x^{6}\sqrt{1-x^{2}}\, dx~~~[3+2] \end{align*}

2023-24 (CBCS)

If \begin{align*} I_{m,n}=\int^{\frac{\pi}{2}}_{0}\sin^{m}{x}\cos^{n}{x} \, dx \end{align*} \(m,n\) being positive integers greater than \(1\), prove that then show that \begin{align*} I_{m,n}=\frac{n-1}{m+n}I_{m,n-2} \end{align*} Hence find the value of \begin{align*} \int^{1}_{0}x^{m}\sqrt{1-x^{2}}\, dx~~~[3+2] \end{align*}
If \(m\) and \(n\) are positive integers, show that \begin{align*} \int^{b}_{a}\big(x-a \big)^{m}\big(b-x \big)^{n}\, dx=\frac{m!n!}{(m+n+1)!}(b-a)^{m+n+1}~~~[4] \end{align*}

2022-23 (CBCS)

No Questons

2021-22 (CBCS)

If \begin{align*} I_{n}=\int^{\frac{\pi}{2}}_{0}\cos^{n-2}{x}\sin{nx} \, dx \end{align*} show that \begin{align*} 2(n-1)I_{n}=1+(n-2)I_{n-1} \end{align*} and hence decuce \begin{align*} I_{n}=\frac{1}{n-1}~~~[5+2] \end{align*}
If \(m\) and \(n\) are positive integers, show that \begin{align*} \int^{b}_{a}\big(x-a \big)^{m}\big(b-x \big)^{n}\, dx=\frac{m!n!}{(m+n+1)!}(b-a)^{m+n+1}~~~[4] \end{align*}
If \begin{align*} I_{n}=\int^{1}_{0}x^{n}\tan^{-1}{x}\, dx,~~n\gt 2 \end{align*} then prove that \begin{align*} (n+1)I_{n}+(n-1)I_{n-2}=\frac{\pi}{2}-\frac{1}{n}~~~[6] \end{align*}
Let \(I_{n}=\int^{1}_{0}(\log{x})n\, dx \). Show that \(I_{n}=(-1)^{n}n!\), \(n\) being positive integer. [2]
Generate a reduction formula for \(\int \tan^{n}{x}\, dx, n\in \mathbb{Z}^{+}\) and \(n\gt 1 \). [2]

2020-21 (CBCS)

If \begin{align*} I_{n}=\int^{1}_{0}x^{n}\tan^{-1}{x}\, dx,~~n\gt 2 \end{align*} then prove that \begin{align*} (n+1)I_{n}+(n-1)I_{n-2}+\frac{1}{n}=\frac{\pi}{2}~~~[4] \end{align*}
Find the reduction formula for \(\int \sin^{m}{x}\cos^{n}{x} \, dx\) where either \(m\) or \(n\) or both are negative integers. And hence find \(\int \frac{\cos^{4}{x}}{\sin^{2}{x}}\, dx\). [6]

2019-20 (CBCS)

If \begin{align*} I_{n}=\int^{\frac{\pi}{2}}_{0}\cos^{n-2}{x}\sin{x} \, dx,~~n\gt 2 \end{align*} Prove that \begin{align*} 2\big(n-1 \big)I_{n}=1+\big(n-2 \big)I_{n-1}~~~[2] \end{align*}
Find the reduction formula for \begin{align*} \int \cos^{m}{x}\sin {nx} \, dx [2] \end{align*}

2018-19 (CBCS)

Obtain the reduction formula for \(\int \cosec^{n} {x}\, dx\). [2]
If \begin{align*} I_{m,n}=\int^{\frac{\pi}{2}}_{0}\sin^{m}{x}\cos^{n}{x} \, dx \end{align*} \(m,n\) being positive integers greater than \(1\), prove that then show that \begin{align*} I_{m,n}=\frac{n-1}{m+n}I_{m,n-2} \end{align*} Hence find the value of \begin{align*} \int^{1}_{0}x^{6}\sqrt{1-x^{2}}\, dx~~~[3+2] \end{align*}

2017-18 (CBCS)

If \begin{align*} I_{n}=\int^{\frac{\pi}{4}}_{0}\tan^{n}{x}\, dx \end{align*} \(n\) being a positive integer greater than \(1\), then prove that \begin{align*} I_{n}=\frac{1}{n-1}I_{n-2}~~~[2] \end{align*}
If \begin{align*} I_{m,n}=\int^{\frac{\pi}{2}}_{0}\cos^{m}{x}\sin {nx} \, dx \end{align*} then show that \begin{align*} I_{m,n}=\frac{1}{m+n}+\frac{1}{m+n}I_{m-1,n-1} \end{align*} also, deduce that \begin{align*} I_{m,n}=\frac{1}{2^{m+1}}\big[2+\frac{2^{2}}{2}+\frac{2^{3}}{3}+…+\frac{2^{m}}{m} \big]~~~[5] \end{align*}

FAQs

What are Reduction Formulae?
Reduction formulae simplify complex integrals by expressing them in terms of simpler integrals using Integral Calculus.
Where are Reduction Formulae applied?
They are widely used in mathematical analysis, physics, and engineering to solve integrals.
How are Reduction Formulae derived?
They are derived using the principles of Differential Calculus and Integral Calculus.
What are some examples of Reduction Formulae?
Examples include formulas for integrals of powers of trigonometric functions and exponential terms.
How do universities test Reduction Formulae?
Universities test these through questions on recursive relationships and practical integrations in Mathematics.
What are the benefits of learning Reduction Formulae?
Learning Reduction Formulae enhances problem-solving efficiency and develops analytical skills in Mathematics.
What are the prerequisites for understanding this topic?
A solid foundation in Differential Calculus and Integral Calculus is required.
How do Reduction Formulae relate to real-world applications?
They are used in engineering and physics to solve integral problems in practical scenarios.
What types of integrals can be simplified using Reduction Formulae?
Reduction Formulae can simplify polynomial, trigonometric, and exponential integrals.
Are Reduction Formulae included in competitive exams?
Yes, many competitive exams feature questions on Reduction Formulae as part of their Mathematics syllabus.