Previous Year Questions on Reduction Formulae
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*}
- 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*}
- No Questons
- 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]
- 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]
- 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*}
- 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*}
- 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.
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