Tricky Integration Problem

$\int _{ \frac { \pi }{ 3 } }^{ \frac { \pi }{ 2 } }{ 64\sin ^{ 2 }{ tcostdt } }$

$u=sint\quad \therefore \quad { u }^{ 2 }=\sin ^{ 2 }{ t } \\ \\ If\quad u=sint,\quad \frac { du }{ dt } =cost,\quad \therefore \quad du=costdt,\quad \therefore \quad \frac { 1 }{ cost } du=dt\\ \\ When\quad t=\frac { \pi }{ 2 } ,\quad u=sin\left( \frac { \pi }{ 2 } \right) =1.\quad When\quad t=\frac { \pi }{ 3 } ,\quad u=sin\left( \frac { \pi }{ 3 } \right) =\frac { \sqrt { 3 } }{ 2 } .\\ \\ So\quad we\quad must\quad integrate:\quad \int _{ \frac { \sqrt { 3 } }{ 2 } }^{ 1 }{ 64{ u }^{ 2 } } \cdot cost\cdot \frac { 1 }{ cost } du,\\ \\ which\quad is:\quad \int _{ \frac { \sqrt { 3 } }{ 2 } }^{ 1 }{ 64{ u }^{ 2 } } du\quad =\quad { \left[ \frac { 64{ u }^{ 3 } }{ 3 } \right] }_{ \frac { \sqrt { 3 } }{ 2 } }^{ 1 }\\ \\ =\left( \frac { 64{ \left( 1 \right) }^{ 3 } }{ 3 } \right) -\left( \frac { 64{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 3 } }{ 3 } \right) =\frac { 64 }{ 3 } -8\sqrt { 3 }$

Tricky Logarithm Problem…

Find the exact solution to the equation ${ 3 }^{ x }{ e }^{ 7x+2 }=15$.

${ 3 }^{ x }{ e }^{ 7x+2 }=15\\ \\ { 3 }^{ x }{ e }^{ 7x }{ e }^{ 2 }=15\\ \\ { 3 }^{ x }{ e }^{ 7x }=\frac { 15 }{ { e }^{ 2 } } \\ \\ { \left( 3{ e }^{ 7 } \right) }^{ x }=\frac { 15 }{ { e }^{ 2 } } \\ \\ \log _{ e }{ \left( { \left( 3{ e }^{ 7 } \right) }^{ x } \right) } =\log _{ e }{ \left( \frac { 15 }{ { e }^{ 2 } } \right) } \\ \\ x\log _{ e }{ \left( 3{ e }^{ 7 } \right) } =\log _{ e }{ \left( \frac { 15 }{ { e }^{ 2 } } \right) } \\ \\ x=\frac { \log _{ e }{ \left( \frac { 15 }{ { e }^{ 2 } } \right) } }{ \log _{ e }{ \left( 3{ e }^{ 7 } \right) } } =\frac { \log _{ e }{ 15-\log _{ e }{ \left( { e }^{ 2 } \right) } } }{ \log _{ e }{ 3+\log _{ e }{ \left( { e }^{ 7 } \right) } } } =\frac { \ln { 15-2\log _{ e }{ \left( e \right) } } }{ \ln { 3+7\log _{ e }{ \left( e \right) } } } \\ \\ =\frac { \ln { 15-2 } }{ \ln { 3+7 } } ,\quad \therefore \quad x=\frac { -2+\ln { 15 } }{ 7+\ln { 3 } } \\$

Ordinary Graph Transformations:

$GRAPH\quad TRANSFORMATIONS:\\ \\ f\left( x+a \right) ,\quad -a\quad x\quad coordinates.\\ \\ f\left( x-a \right) ,\quad +a\quad x\quad coordinates.\\ \\ f(ax),\quad Multiply\quad x\quad coordinates\quad by\quad \frac { 1 }{ a } .\\ \\ f\left( x \right) +a,\quad +a\quad y\quad coordinates.\\ \\ f\left( x \right) -a,\quad -a\quad y\quad coordinates.\\ \\ af\left( x \right) ,\quad Multiply\quad y\quad coordinates\quad by\quad a.$

(a+b)(a-b)=a^2-b^2 (The Real Proof) (Difference Of Two Squares Demystified)

Ever wondered why (a+b)(a-b)=a^2-b^2? The videos below will reveal to you why we accept this fact. Enjoy!

1) Why (a+b)(a+b)=a^2+2ab+b^2

2) Why (a+b)(a-b)=a^2-b^2

How To Come Up With The Quadratic Formula – From Scratch

Learn how to come up with the quadratic formula from scratch. Watch the video below…

Trapezium Rule Formula – Derivation

Find out how to come up with the Trapezium Rule formula from scratch.

1) Derive the formula for the area of trapeziums:

2) Use the area of trapeziums formula to come up with the Trapezium Rule formula:

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Derive the formula to find areas underneath curves