I did help with this problem i did half of it and i need help with other half.Please help me fill in the blanks **Integration by Parts**Apply the rule of integration by parts:\[ \int u \, dv = uv - \int \boxed{} \, du \]1. Let \( dv = e^w \, dw \) and integrate the differential equation to obtain the function \( v \).\[ v = \int dv = \int e^w \, dw = \boxed{} \]2. And let \( u = \cos w \). Differentiate \( u \) in terms of \( w \).\[ du = \boxed{} \, dw \] ### Evaluating the Integral Using Substitution and Integration by PartsWe aim to evaluate the integral \[ \int \cos(\ln x) \, dx \]using substitution first, followed by integration by parts.#### Step 1: SubstitutionTo find the integral \(\int \cos(\ln x) \, dx \), we will perform a substitution and then use the formula for integration by parts.1. Let \( w = \ln x \). We need the derivative of \( w \) in terms of \( x \).\[ dw = \frac{1}{x} \, dx \]2. From this, we can express \( dx \):\[ dx = e^w \, dw \]3. Substituting \( w \) and \( dx \), the integral becomes:\[ \int \cos(\ln x) \, dx = \int \cos(w) \, e^w \, dw \]We are now prepared to proceed with integration by parts in the next steps.

Here we will fill all blacks of prat 2 as shown below

Apply the rule of integration by parts. -S Sudv. dv = uv - du = Let dv= ew dw and integrate the differential equation to obtain the function v. v=[ov = [H ² ON = [ Se V ew dw And let u cos w. Differentiate u in terms of w. du dw

Apply the rule of integration by parts. -S Sudv. dv = uv - du = Let dv= ew dw and integrate the differential equation to obtain the function v. v=[ov = [H ² ON = [ Se V ew dw And let u cos w. Differentiate u in terms of w. du dw

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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