log[secx ·(d y)/(d x)]=sinx+2 logsinx . Then y((π)/(2)) is (a) 0 (b) 1 (c) e (d) (1)/(e)

Name: Problem 117, Chapter 3: IIT JEE Super Course in Mathematics: Calculus
Uploaded: 2021-07-17T05:38:23-08:00
Duration: 6 min 2 s
Channel: Urvashi Arora
Description: log[secx ·(d y)/(d x)]=sinx+2 logsinx . Then y((π)/(2)) is (a) 0 (b) 1 (c) e (d) (1)/(e)

step 1 Hello. The question that is given to us is log of second. That is basically from the integral ca

Log[secx ·(d y)/(d x)]=sinx+2 logsinx . Then y((π)/(2)) is...

Key Concepts

Properties of Logarithms

This concept involves using the laws of logarithms such as turning a sum of logarithms or logarithms with multipliers into products or powers. In the context of the problem, these properties allow for rewriting and simplifying expressions before further manipulation, for example by converting logarithmic sums into the logarithm of a product.

Exponentiation to Eliminate Logarithms

Once the logarithmic equation is simplified, exponentiation is applied to both sides to remove the logarithm and obtain an equation in a more tractable exponential form. This step is essential in moving from a logarithmic expression to an algebraic or differential equation that can be solved directly.

Differential Equations

The problem involves a differential equation where the derivative of the function is given in terms of the independent variable and possibly the function itself. Understanding how to set up and solve differential equations is crucial to finding the unknown function, particularly when it involves techniques like separating variables.

Integration Techniques

After isolating the derivative, integration is used to recover the original function from its derivative. This often requires using specific integration techniques to handle functions that might not be directly integrable without a substitution or transformation.

Substitution Method in Integration

When faced with integrals that contain composite functions or complicated expressions, the substitution method (or u-substitution) helps simplify the integral into a more standard form. In this context, it often transforms an integral into one that can be directly integrated using known formulas.

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