Find the indefinite integral. ∫−11−(4t+1)2dt

Question

Find the indefinite integral.  ∫−11−(4t+1)2dt

Wendy Boykin · Accepted Answer

Step 1  We have the given integral,  ∫−11−(4t+1)2dt  By substituting u=4t+1, that implies du=4dt  Therefore, the integral becomes,  =∫−11−u2(14)du  =−14∫11−u2du  Step 2  Now, by substituting sin⁡x=u, that implies, cos⁡xdx=du  Therefore, the integral becomes,  =−14∫11−sin2x(cos⁡x)dx  =−14∫1cos2x(cos⁡x)dx  =−14∫1cos⁡x(cos⁡x)dx  =−14∫dx  =−14x+C  Step 3  But we know, sin⁡x=u, therefore x=sin−1(u)  By substituting this value of x,  =−14sin−1u+C  Then by resubstituting u=4t+1,  =−14sin−1(4t+1)+C

Tiefdruckot · Accepted Answer

∫−11−(4t+1)2dt  ∫−141−u2du  −14⋅∫11−u2du  −14⋅arcsin⁡(u)  −14⋅arcsin⁡(4t+1)  −arcsin⁡(4t+1)4  Add C  Solution:  −arcsin⁡(4t+1)4+C

nick1337 · Accepted Answer

Step 1 Given: ∫−11−(4t+1)2dt Substitution u=4t+1⇒dudt=4 =−14∫11−u2du This is the well-known tabular integral: =arcsin⁡(u) We substitute the already calculated integrals: −14∫11−u2du =−arcsin⁡(u)4 Reverse replacement u=4t+1: =−arcsin⁡(4t+1)4 Answer: =−arcsin⁡(4t+1)4+C

Find the indefinite integral. \int \frac{-1}{\sqrt{1-(4t+1)^{2}}}dt

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