sbhatnagar said:\( \displaystyle \int \sin^{12}(7x) \cos^{3}(7x) \ dx = \int \sin^{13}(7x) \{ 1-\sin^2(7x)\}\cos(7x) \ dx\)
Now substitute \( u=\sin(7x) \).
\( \displaystyle \int \sin^{13}(7x) \{ 1-\sin^2(7x)\}\cos(7x) \ dx = \frac{1}{7}\int u^{12}(1-u^2) \ du\)
Can you take it from here?
An integral is a mathematical concept that represents the area under a curve in a graph. It is a fundamental concept in calculus and is used to solve problems related to rates of change and accumulation.
An integral can be considered "hard" if it is difficult to solve using standard techniques, such as substitution or integration by parts. This can be due to the complexity of the integrand or the limits of integration, among other factors.
Integrals are important in science because they allow us to model and analyze real-world phenomena, such as motion, population growth, and chemical reactions. They also provide a way to calculate important quantities like area, volume, and average values.
Some strategies for solving a hard integral include using techniques like u-substitution, integration by parts, and trigonometric substitutions. It can also be helpful to break down the integral into smaller, more manageable parts and to use computer software or numerical methods when necessary.
Integrals have many real-world applications, including calculating the area under a pressure-volume curve to determine the work done by a gas, finding the average velocity of an object over a period of time, and determining the concentration of a substance in a solution over time in a chemical reaction.